The Binomial Theoremweb2.hunterspt-h.schools.nsw.edu.au/studentshared...Binomial product: The...
Transcript of The Binomial Theoremweb2.hunterspt-h.schools.nsw.edu.au/studentshared...Binomial product: The...
Binomial product: The product of two or more expressions each consisting of two terms e.g. ( )a x n+
Choose notation (or combination): Describes the number of ways of choosing r items from n in no particular order
Coefficient: A constant multiplied by a pronumeral e.g. 3x2 has coefficient 3
TERMINOLOGY
The Binomial Theorem
10
467Chapter 10 The Binomial Theorem
DID YOU KNOW?
The binomial theorem was studied by mathematicians from very early times. For example, Euclid discovered the special expansion for ( )x y 2+ around 300 BC. Euclid is mainly famous for his work on geometry, but he also studied arithmetic and number theory. He founded the first school of mathematics, in Alexandria in Egypt.
Omar Khayyam, working around 1100 AD, discovered the expansions of ( ) ( )x y , x y4 5+ + and .( )x y 6+ He is mainly remembered as a poet (he wrote the Rubaiyat), but he was also an astronomer
and a mathematician. He wrote a book about algebra.Blaise Pascal (1623–62) is associated with the binomial theorem because of his famous ‘Pascal’s
triangle’, which is related to the coefficients of binomials. However, he approached the binomial theorem from the standpoint of probability rather than algebra.
Sir Isaac Newton (1642–1727) discovered the general rules for coefficients of binomial expansions, which Leonhard Euler (1707–83) later proved.
INTRODUCTION
You have alreadY studied binomial products. a special result of these products is .( )x y 2+ the binomial theorem looks at the expansion of the more general binomial ( )x y n+ .
the binomial theorem is associated with algebra and probability. in this chapter, you will study some properties of binomial products and their coefficients. in the next chapter you will see how the binomial theorem is used in probability theory.
Class Investigation
Can you complete the next 3 lines of Pascal’s triangle? is the triangle 1. always symmetrical?
1
1
3
1 1
3
1
4 4
1
1
1
1
2
6
Find the sum of each line.2. e.g. 1 3 3 1 8+ + + =Can you find a pattern for these sums? What would be the sum of the numbers in the nth line?
Combinations
468 Maths In Focus Mathematics Extension 1 HSC Course
in the Preliminary Course, you studied factorial notation and combinations. We revise these here.
Factorial notation:
! ( ) ( ) ( )… · ·n n n n n1 2 3 3 2 1= − − −
0! 1=
( )! !!C
n r rnn
r =−
n( )! !
!n r r
n=−ra k
Combinations:
We can also use choose notation:
ExamplEs
evaluate
1. 4!
Solution
4! 4 3 2 1
24
==
2. C107
Solution
( )! !!
! !!
C10 7 7
10
3 710
107 =
−
=
......
3 2 1 7 6 5 3 2 110 9 8 7 3 2 1
120
=
=
You can also evaluate these questions on a calculator.
´ ´ ´
´ ´ ´ ´ ´ ´´ ´ ´ ´ ´
´ ´ ´´ ´
469Chapter 10 The Binomial Theorem
3. 51a k
Solution
( )! !!
! !!
5 1 15
4 15
4 3 2 1 15 4 3 2 1
5
5 =−
=
=
=
1a k
4. C60
Solution
( )! !!
! !!
C6 0 0
6
6 06
6 5 4 3 2 1 16 5 4 3 2 1
1
60 =
−
=
=
=
5. 88a k
Solution
( )! !!
! !!
8 8 88
0 88
1 8 7 6 5 4 3 2 18 7 6 5 4 3 2 1
1
8 =−
=
=
=
8a k
Properties of coefficients
C 1n0 =
´ ´ ´ ´´ ´ ´ ´
´ ´ ´ ´ ´
´ ´ ´ ´ ´´ ´ ´ ´ ´2 ´
´ ´ ´ ´ ´ ´ ´ ´´ ´ ´ ´ ´ ´ ´=
470 Maths In Focus Mathematics Extension 1 HSC Course
Proof
( )
( )
a b C a C a b C a b C a b C ba b
C C C C C
C
1 0
1 0 1 1 0 1 0 1 0 0
1
Let and
n n n n n n n n n nn
n
n n n n n n n n n nn
n
n
0 11
22 2
33 3
0 11
22 2
33 3
0
f
f
+ = + + + + += =
+ = + + + + +=
− − −
− − −
C 1nn =
Proof
( )
( )
a b C a C a b C a b C a b C ba b
C C C C C
C
C C
0 1
0 1 0 0 1 0 1 0 1 1
1
1
Let and
n n n n n n n n n nn
n
n n n n n n n n n nn
n
nn
n nn
0 11
22 2
33 3
0 11
22 2
33 3
0`
f
f
+ = + + + + += =
+ = + + + + +== =
− − −
− − −
in general,
C Cnk
nn k= −
C C Cnk
nk
nk
11
1= +−−
− for ≤ ≤k n1 1−
Proof
(1 )
(1 )
( ) ( )
( ) ( )
( )
( ) ( )
( )
x C C x C x C x
x x C x C x C x
C x C x
x x x
C C x C x C x
C x C x C x C x C x
C C C x C C x
C C x C x
x x
x
C C x C x C x C x C x
1 1
1 1
1
LHS
RHS
n n n n nn
n
n n n n
nn
n nn
n
n n
n n n nn
n
n n n nn
n nn
n
n n n n n
nn
nn
n nn
n
n
n
n n n n nk
k nn
n
1 10
11
12
2 11
1
1 10
11
2 12
3
12
1 11
1 1
10
11
12
2 11
1
10
11
2 12
3 12
1 11
10
11
10
12
11
2
11
12
1 11
1
0 1 22
33
`
`
f
f
f
f
f
f f
+ = + + + ++ = + + +
+ ++ + +
= + + + ++ + + + + +
= + + + + ++ + +
= + += += + + + + + + +
− − − − −−
−
− − − −
−−
− −−
− −
− − − −−
−
− − − −−
− −−
− − − − −
−−
−−
− −−
−
471Chapter 10 The Binomial Theorem
equating coefficients gives
,
, ,
C C C
C C C
C C C
n n n
n n n
nn
nn
nn
11
11
0
21
21
1
11
11
2
f
= += += +
− −
− −
−−
−−
−
in general, .C C Cnk
nk
nk
11
1= +−−
−
ExamplEs
1. evaluate .C99
Solution
C 1=99
2. show that .C C38
5=8
Solution
C8
( )! !!
! !!
( )! !!
! !!
C
C
C
8 3 38
5 38
8 5 58
3 58
So
83
85
38
5
=−
=
=−
=
=
3. show that 7 6 6= +4 3 4a a ak k k
Solution
( )! !!
! !!
( )! !!
( )! !!
! !!
! !!
7 4 47
3 47
6 3 36
6 4 46
3 36
2 46
7
6 6
LHS
RHS
=
=−
=
= +
=−
+−
= +
4
3 4
a
a a
k
k k
CONTINUED
472 Maths In Focus Mathematics Extension 1 HSC Course
! !!
! !!
! !!
! !!
! !( !) ( !)
! !( !)
! !!
LHS
3 3 46 4
2 4 36 3
4 36 4
4 36 3
4 34 6 3 6
4 37 6
4 37
7 6 6So
= +
= +
=+
=
=
=
= +4 3 4a a ak k k
1. evaluate7!(a) 8!(b) 3!(c) 5!(d) 0!(e)
(f) ! !10 2−3(4!)(g) 5(6!)(h)
(i) 59!
(j) 4!8!
2. evaluate(a) C4
6
(b) C102
(c) 96a k
(d) C811
(e) C130
(f) 55a k
(g) C14
(h) 125a k
(i) C12
(j) 108a k
3. show that
(a) 9 9=5 4a ak k
(b) C C=72 5
7
(c) 12 12=5 7a ak k
(d) C C=113 8
11
(e) C C=101 9
10
(f) 9 8 8= +7 6 7a a ak k k
(g) 11 10 10= +6 5 6a a ak k k
(h) C C C= +75 4 5
6 6
(i) 10 9 9= +6 5 6a a ak k k
(j) 7 6 6= +3 2 3a a ak k k
4. show that nn =k n k−a ak k.
5. evaluate x if C .C =x 277
6. if ,12 12=y3
a bk l evaluate y.
7. Find the value of a if C C= .a11
811
8. solve n 10 10= +6 5 6a a ak k k.
9. solve .CC +C = k7 720 19 19
10.1 exercises
´´
´´
´ ´
473Chapter 10 The Binomial Theorem
10. Factorise(a) ! ( ) !k k 1+ +(b) ! ( ) !r r 1+−(c) ( ) ! !n n1+ +(d) ( ) ! ( ) !k k1 1+ +−(e) ( ) ! ( ) !p p1 1+ − −(f) ! ( ) ! ( ) !t t t1 1+ + − −
11. simplify
(a) !
( ) !k
k 1−
(b) ( ) !( ) !kk
11+
−
(c) ( ) !( ) !nn
21
−−
(d) ( 1) !
!m
m+
(e) ( 2) !( 1) !kk
+−
12. simplify and leave your answer in factorial form.
(a) C
!6
83
(b) C
!85
9
(c) C
!54
6
(d) C
!74
9
(e) C10
!87
13. simplify
(a) 7
7
3
4
a
a
k
k
(b) C
C
5
116
11
(c) C
C12
3
412
(d) C
C9
6
97
(e) 14 148 7a ak k
(f) C
C6
63
4
(g) C
C8
5
86
14. simplify C
Cn
.nk
k
1−
15. show that .n nk
n11
1= +−−
−k ka b bk l l
Class Discussion
Combinations and Pascal’s triangle are closely related as Pascal’s triangle can be written in terms of combinations.
0C0
3C3
2C2
1C11C0
2C12C0
4C0
3C23C0
3C1
4C14C4
4C34C2
Check these values to show this is true. Can you write the next 1. 3 lines?Can you see from Pascal’s triangle that 2. ?C Cn n
k n k= −
Can you see from Pascal’s triangle that 3. ?CC +C =nk
nk
nk
11
1−−
−
¸
474 Maths In Focus Mathematics Extension 1 HSC Course
Investigation
expand these binomial products.1. (a) 0( )x1 +(b) 1( )x1 +(c) ( )x1 2+(d) ( )x1 3+(e) ( )x1 4+By looking at these expansions and Pascal’s triangle, could you 2. predict the expansion of ?( )x1 5+ What about ?( )x1 6+For each expansion, find the sum of the coefficients. What do you 3. notice? Could you predict the sum of coefficients of ( )x1 5+ or
?6( )x1 +
the binomial expansions and Pascal’s triangle are related.
ExamplEs
1.
expand (a) .( )x1 2+Find the sum of coefficients of the expansion.(b) how many terms are in the expansion?(c)
Solution
(a) .( )x x x1 1 22 2+ = + +sum of the coefficients:(b)
1 2 1 4
22
+ + ==
Notice that these coefficients form a row in Pascal’s triangle.there are 3 terms in the expansion.(c)
2.
expand (a) .( )x1 3+Find the sum of coefficients of the expansion.(b) how many terms are in the expansion?(c)
We can use the expansion .(a b) a 2ab b22+ + +=2
Binomial Theorem
You have already studied binomial products in the form of ( ) ( )x a y b+ + as well as special products such as perfect squares.
Expansion of (1 )x n+
The coefficient is the constant in each term.
475Chapter 10 The Binomial Theorem
Solution
(a) 2( ) ( ) ( )
( ) ( )
x x x
x x x
x x x x x
x x x
1 1 1
1 1 2
1 2 2
1 3 3
3
2
2 2 3
2 3
+ = + += + + += + + + + += + + +
(b) sum of the coefficients:
1 3 3 1 8
23
+ + + =
= Notice that the coefficients form a row in Pascal’s triangle.
there are 4 terms in the expansion.(c)
in general, the binomial expansion of ( )x1 n+ has coefficients that form the rows of Pascal’s triangle.
the sum of the coefficients of ( )x1 n+ is 2n
there are n 1+ terms in the binomial expansion of n( )x1 +
since Pascal’s triangle can also be written using the combinations Cnk, we can
write the binomial expansion as
xC C C x C x C x C xC xn n n(1 )x
x x x x xn n n n nk
nn0 1 2 3
n n nk
k nn
n nn
n
k n
0 1 22
33
11
2 3
f f
f f
+ = + + + + + + + +
= + + + + + + +
−−
a a a a a ak k k k k k
We can also write this binomial expansion in sigma notation.
C xn( )x
xnk
1 nk
k
k
n
k
k
n0
0
+ =
=
=
=a k
/
/
the ( )k 1 th+ term of the binomial expansion is Cnk x
k
476 Maths In Focus Mathematics Extension 1 HSC Course
ExamplEs
1.
expand the binomial product (a) .6( )x1 +Find the sum of its coefficients.(b) Write the binomial expansion in sigma notation.(c)
Solution
(a) C C x C x C x C x C x C x6( )x
x x x x x x
1
1 6 15 20 15 6
6 61
62
2 63
3 64
4 65
5 66
6
2 3 4 5 6
+ = + + + + + += + + + + + +
0
sum of coefficients of (b) ( )x1 n+ is 2n
so the sum of coefficients of 6( )x1 + is 26 or 64.
(c) C xn( )x1 nk
k
k
n
0+ =
=/
so C x6( )x1 kk
k
6
0
6
+ ==/
2. evaluate the coefficient of x9 in the binomial expansion of .( )x1 12+
Solution
the general term of a binomial expansion is nCkxk.
so in the expansion of ( )x1 12+ the term involving x9 is 12C9x9.
the coefficient is .C 220=129
3. how many terms are in the expansion of ?( )x1 20+
Solution
there are n 1+ terms in the expansion of .( )x1 n+ so there are 20 1+ or 21 terms in the expansion of .20( )x1 +
4. expand .x5 k
k 0
5
= ka k/
Solution
x x x x x x
x x x x x1 5 10 10 5
5 5 5 5 5 5 5k
k 0
52 3 4 5
2 3 4 5
= + + + + +
= + + + + += k 0 1 2 3 4 5a a a a a a ak k k k k k k/
You could simply add the coefficients from part (a).
This is the 10th term since
the 1st term is C x0
012.
477Chapter 10 The Binomial Theorem
5. Find the 4th term of the binomial product .( )x1 + 9
Solution
the ( )k 1 th+ term of ( )x1 n+ is Cnkx
k.For the 4th term:k
k
1 4
3
+ ==
so the 4th term of ( )x1 9+ is .C x x843
3 3=9
1. (i) expand each binomial product.
Find the sum of coefficients.(ii) Find the number of terms.(iii) Write each product in sigma (iv)
notation.(a) ( )x1 3+(b) ( )x1 4+(c) ( )x1 7+(d) ( )x1 6+(e) ( )x1 5+
2. Find the sum of coefficients of each (i)
binomial product in index formthe number of terms(ii) (a) ( )x1 25+(b) ( )x1 34+(c) ( )x1 17+(d) ( )x1 63+(e) 40( )x1 +
3. (i) Write as a binomial product and
write out the expansion of (ii)
(a) C xkk
k
4
0
4
=/
(b) x7 k
k 0
7
= ka k/
(c) C x3k
k
k 0
3
=/
(d) C xkk
k
6
0
6
=/
(e) x8 k
k 0
8
= ka k/
4. Find the coefficient of x4 in the expansion of(a) ( )x1 7+(b) 9( )x1 +
(c) C xkk
k
6
0
6
=/
(d) ( )x1 8+(e) ( )x1 5+
5. Find the coefficient of x3 in the expansion of(a) ( )x1 9+
(b) C x5k
k
k 0
5
=/
(c) ( )x1 11+(d) ( )x1 10+
(e) x4 k
k 0
4
= ka k/
6. Find the coefficient of x2 in the expansion of (a) 3( )x1 +
(b) C xkk
k
6
0
6
=/
(c) ( )x1 10+(d) ( )x1 7+(e) ( )x1 14+
10.2 exercises
478 Maths In Focus Mathematics Extension 1 HSC Course
7. Find the coefficient of x5 in the expansion of(a) ( )x1 10+(b) ( )x1 12+(c) ( )x1 8+(d) ( )x1 15+
(e) x7 k
k 0
7
= ka k/
8. Find the coefficient of the 5th term of(a) ( )x1 5+(b) ( )x1 7+
(c) x8 k
k 0
8
= ka k/
(d) 9( )x1 +(e) ( )x1 11+
9. Find the 3rd term of(a) ( )x1 8+(b) ( )x1 3+(c) ( )x1 4+(d) ( )x1 7+
(e) x6 k
k 0
6
= ka k/
10. Find the kth term of(a) ( )x1 n 1+ +
(b) ( )x1 n2+
(c) xn 1 k
k
n
0
1 −=
−
kb l/
(d) ( )x1 n2 1+ +
(e) ( )x1 n3 1+ −
Remember the general term is the (k 1)th+ term so the 5th term is .C xn
44
Expansion of ( )a x n+
Investigation
expand these binomial products.1. (a) ( )a x 0+(b) ( )a x 1+(c) ( )a x 2+(d) ( )a x 3+(e) ( )a x 4+By looking at these expansions and Pascal’s triangle, could you 2. predict the expansion of ?( )a x 5+ What about ?( )a x 6+For each expansion, find the sum of the coefficients. What do you 3. notice? Could you predict the sum of coefficients of 5( )a x+ or
?( )a x 6+
the binomial expansion of ( )a x n+ and Pascal’s triangle are related in the same way as ( )x1 n+ .
the sum of the coefficients Cnk of ( )a x 2isn n+
there are n 1+ terms in the binomial expansion of ( )a x n+
479Chapter 10 The Binomial Theorem
We can write the binomial expansion as
C a C a x C a x C a x
C a x C x
n n( )a x
a a x a x a x a x xn n n n nk
nn0 1 2 3
n n n n n n n
nk
n k k nn
n
n n n n n k k n
0 11
22 2
33 3
1 2 2 3 3
f
f
f f
+ = + + + ++ + +
= + + + + + + +
− − −
−
− − − −a a a a a ak k k k k k
We can also write this binomial expansion in sigma notation.
C a xn n( )a x
a xnk
kn k k
k
n
n k k
k
n0
0
+ =
=
−
=
−
=a k
/
/
Notice that this is the same as n( )a x a xnk
k n k
k
n
0+ = −
=a k/ since n
k n k= −
na ak k. Can you
see why?
the ( )k 1 th+ term of the binomial expansion is C a xnk
n k k−
ExamplEs
1. expand .( )x4 5+
Solution
C C x C x C x C x C x4 4 4 4 45( )
( ) ( ) ( ) ( ) ( ) ( )
x
x x x x x
x x x x x
4
1 1024 5 256 10 64 10 16 5 4 1
1024 1280 640 160 20
50
5 51
4 52
3 2 53
2 3 54
4 55
5
2 3 4 5
2 3 4 5
+ = + + + + += + + + + += + + + + +
2. expand .( )x3 2 4+
Solution
C( ) ( ) ( ) ( ) ( ) ( )( ) ( )C x C x C x x C3 3 2 3 2 3 2 22 34( )
( ) ( )( ) ( )( ) ( )( ) ( )
x
x x x x
x x x x
3 2
1 81 4 27 2 6 9 4 4 3 8 1 16
81 216 216 96 16
4 40
4 41
3 42
23
44
4
4 3 2
4 3 2
+ = + + + += + + + += + + + +
CONTINUED
480 Maths In Focus Mathematics Extension 1 HSC Course
3. expand .( ) ( )C x y2 −3k
k k
k
3
0
3−
=/
Solution
( ) ( )
( ) ( ) ( ) ( )( ) ( )
C x y
C x C x y C x y C y
2
2 2 2
−
− − −23
3
2 33
( )
( ) ( ) ( ) ( )( ) ( )
x y
x x y x y y
x x y xy y
2
1 8 3 4 3 2 1
8 12 6
kk k
k
3 3
0
33
30
31 2 3
3
3 2 2
3 2 2 3
= −
= + + += + − + − + −= − + −
−
=/
4. if ( ) a b1 3 34+ = + , evaluate a and b.
Solution
expand ( )1 3 4+
( )C( ) ( ) ( )C C C C1 1 1 1 +4 2 4( )
( ) ( ) ( )
1 3 3 3 3 3
1 4 3 6 3 4 3 3
1 4 3 6 9 4 27 81
1 4 3 6 3 4 9 3 9
1 4 3 18 4 3 3 9
28 16 3
40
4 41
3 42
2 43
1 3 44
2 3 4
+ = + + += + + + += + + + += + + + += + + + += +
so a 28= and b 16= .
5. By writing 0.999 as . ,1 0 001− evaluate 0.9995 correct to 3 decimal places.
Solution
5
( . ) ( . ) ( . )
1( 0.001) ( 0.001)
C C C C
C C
1 1 0 001 1 0 001 1 0 001+ − − −− −
2
4
3
. ( . )
( . ) ( . ) ( . ) ( . )
( . ).
0 999 1 0 001
1 5 0 001 10 0 001 10 0 001 5 0 001
0 0010 995
5 5
50
5 51
4 52
3 53
2 3
54
55
2 4
5
= −= + +
+ += + − + − + − + −
+ −=
6. Find the 3rd term of the expansion of 6( )x2 3− .
Solution
the ( )k 1 th+ term of ( )a x n+ is C a x .nk
n k k−
For the 3rd term: k
k
1 3
2
+ ==
´ ´´
481Chapter 10 The Binomial Theorem
so the 3rd term of 6( )x2 3− is( ) ( )C x2 3− 26
26 2−
( ) ( )
( )
x
x
x
15 2 9
135 16
2160
4
4
4
===
1. Write each product in sigma notation.(a) ( )x6 5 12+(b) ( )a b 18−(c) ( )y3 2 24+
16( )x y2−(d) ( )x y5 7 12+(e)
2. expand each binomial product.(a) ( )a x 4+(b) ( )a x 6+(c) ( )a x 5+(d) ( )a2 1 3+(e) ( )x 2 7−(f) ( )x4 32 4+(g) ( )x3 2 6−(h) ( )a b4 5 3−(i) ( )m2 3 5+(j) ( )x1 2 8−
3. (i) Write as a binomial product and
find the expansion of(ii)
(a) C a xkk k
k
3 3
0
3−
=/
(b) ( )2−( )x35 k k
k
5
0
5−
= ka k/
(c) ( )C x y2 kk
k
k
6 6
0
6−
=/
(d) ( )b5( )C a2 kk
k
k
4 4
0
4−
=/
(e) 7 x yk k
k
7
0
7 −
= ka k/
(f) ( )q3−( )C p4 k3 − kk
k
3
0
3
=/
(g) ( )n2( )C 3 kk
k
k
5 5
0
5−
=/
(h) ( )b−( )C a2 kk
k
k
6 6
0
6−
=/
(i) ( )c4−( )ab34 k k
k
4
0
4−
= ka k/
(j) ( )x−( )27 k k
k
7
0
7−
= ka k/
4. Write as a binomial product without expanding.
(a) ( )7( )C a k10k
k
k
10
0
10−
=/
(b) )y(−( )69 k k
k
9
0
9−
= ka k/
(c) ( )b4( )C a3kk k
k
7 7
0
7−
=/
(d) ( )3( )C xkk k
k
8 2 8
0
8−
=/
(e) ( )q−( )C p5kk k
k
11 11
0
11−
=/
(f) ( )9−( )xn 4 n k k
k
n
0
−
= ka k/
(g) ( )b( )C a3nk
n k k
k
n2
0
−
=/
(h) ( )2−k
( )xn 31 n k k
k
n1
0
1 + + −
=
+
b l/
(i) ( )b6( )C an 1−k
n k k
k
n2 1
0
1− −
=
−
/
(j) ( )y7( )C x8nk
n k k
k
n2 2 2
0
2−
=/
10.3 exercises
482 Maths In Focus Mathematics Extension 1 HSC Course
5. expand(a) ( )2 1 5+(b) ( )3 1 6−
(c) k3 −( ) ( )2 2 33k
k
0
3
−= ka k/
(d) ( )3 5 4+
(e) 5 ( ) ( )C 3 2kk
k k
0
55 −
=
−/
(f) x32
4
+b l
(g) x x1 5
+b l
(h) x12
3
−b l
(i) x x6 2k
k
k
6
0
6−
= ka bk l/
(j) a b3 2
3
−c m
6. evaluate a and b if ( ) a b2 3 23+ = + .
7. if ( ) a b2 5 54− = + , evaluate a and b.
8. if ( ) ( ) a b3 2 2 26k
k k
0
66 − = +
=
−
ka k/ ,
evaluate a and b.
9. evaluate x and y if ( ) x y3 1 5− = + .
10. if ( ) ( ) p q4 24k
k k
0
44 = +
=
−
ka k/ ,
evaluate p and q.
11. if ( ) a b2 3 2 33+ = + , evaluate a and b.
12. By writing 0.99 as . ,1 0 01− find to 4 decimal places.
0.99(a) 3
0.99(b) 4
0.99(c) 5
0.99(d) 6
0.99(e) 7
13. By writing 0.98 as . ,1 0 02− evaluate to 4 decimal places.
0.98(a) 2
0.98(b) 5
0.98(c) 3
0.98(d) 4
0.98(e) 6
14. By writing 1.01 as . ,1 0 01+ evaluate to 4 decimal places.
1.01(a) 3
1.01(b) 5
1.01(c) 4
1.01(d) 7
1.01(e) 6
15. Find the coefficient of the 3rd term in the expansion of(a) ( )x3 6+
(b) C x 2kk k
k
5 5
0
5−
=/
(c) ( )a5 7−(d) ( )n3 2 9+
(e) ( ) ( )C t2 3−kk k
k
10 10
0
10−
=/
16. Find the coefficient of the 5th term in the expansion of(a) ( )x y2 8+
(b) ( ) ( )a2 17 k k
k
7
0
7
−−
= ka k/
(c) ( ) ( )a b7 45 k k
k
5
0
5
−−
= ka k/
(d) ( )n2 3 9−
(e) ( ) ( )C x 3kk k
k
6 2 6
0
6−
=/
17. Find the coefficient of the 6th term (in coefficient and index form) in the expansion of(a) ( )a6 15+
(b) ( ) ( )C x5 4kk k
k
10
0
10
−−
=
10/(c) ( )a b2 3 14+(d) ( )x3 5 9−
(e) ( ) ( )t q20 7 5k k
k
20
0
20
−−
= ka k/
18. Find the ( 1)k th+ term in the expansion of(a) ( )x 5 8+(b) ( )a2 3 12+(c) ( )a b5 6−(d) ( )x y4 15−(e) ( )a b3 2 21−
483Chapter 10 The Binomial Theorem
19. Find the kth term in the expansion of (a) ( )x2 9+
(b) ( ) ( )C x y5 2kk k
k
5 5
0
5−
=/
(c) ( ) ( )d3 28 k k
k
8
0
8
−−
= ka k/
(d) 13( )m n6−(e) ( )a b3 2 20−
20. Find the ( )k 1 th− term in the expansion of
(a) ( )x3 6+
(b) C a 2kk k
k
9 9
0
9−
=/
(c) ( )a5 3 6+(d) ( )x3 4 18−
(e) ( ) ( )C x y3 2−8k
k k
k
2 8
0
8−
=/
21. Write out the expansion of ( )sin x1 5+ .
22. simplify ( ) ( )2 1 2 14 4+ + − .
ExamplEs
1.
(a) Find the general ( )k 1 th+ term of the binomial product .( )x3 22 6−(b) Find the coefficient of x4 in the expansion of .( )x3 22 6−
Solution
the (a) ( )k 1 th+ term is ( ) ( )C x3 2− .kk k6 2 6 −
(b) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
C x
C x
C x
3 2
3 2
3 2
−−−
kk k
kk k k
kk k k
6 2 6
6 6 2 6
6 6 12 2
==
−
− −
− −
For the term involving x4:x xk12 2 4=−
so k
k
k
k
12 2 4
12 2 4
8 2
4
− == +==
When 4,k= we have the 5th term of the expansion ( 1)k + .
( ) ( ) ( )C x3 2−( ) ( ) ( )C x3 2−6 4
2
− 4T ( )5
64
12 2 4
64
4 4
==
−
the coefficient is ( ) ( )C 3 2 2160− =464
2 .
CONTINUED
Further Work with Coefficients
once you are familiar with binomial products and their expansions, you can do further work with coefficients.
484 Maths In Focus Mathematics Extension 1 HSC Course
2. Find the constant term of .xx22
2
8
+c m
Solution
the ( 1)k th+ term is ( )xx
8 2kk
2 82
−
ka ck m
xx
xx
x x
x
2
2 1
2
2
8
8
8
8
kk
k
k kk
k k k
k k
16 22
16 22
16 2 2
16 4
=
=
=
=
−
−
− −
−
k
k
k
k
a e
a e
a
a
k o
k o
k
k
the constant term does not involve x, so we look for the term with x0:x xk16 4 0=−
so k
k
k
16 4 0
16 4
4
− ===
When 4,k= we have the 5th term of the expansion ( 1)k +
T x
x
8 2
8 2
8 2
1120
( )5
4 16 4 4
4 0
4
=
=
=
=
−
4
4
4
a
a
a
k
k
k
3. Find the coefficient of x3 in the expansion of .xx
2 52
9
−c m
Solution
( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
T C a b
C xx
C xx
C x
C x
2 5
2 5
2 5
2 5
kn
kn k k
kk
k
kk k
k
k
kk k k k
kk k k
1
9 92
9 9 92
9 9 9 2
9 9 9 3
=
= −
= −
= −= −
+−
−
− −
− − −
− −
c
e
m
o
For the term involving ,x3
( ) ( ) ( )
( ) ( )
x xk
k
k
T C x
x
x
9 3 3
6 3
2
2 5
36 128 25
115 200
( )
k9 3 3
2 19
29 2 2 9 3 2
3
3
`
`
=− =
==
= −==
−
+− −
the coefficient of x3 is 115 200.
485Chapter 10 The Binomial Theorem
Greatest coefficient
You might have noticed that the coefficients often increase in size and then decrease. We can find the greatest coefficient without having to expand the binomial product.
in order to do this, we need to compare the (k + 1)th and kth terms.
in the expansion of ( ) ,a b n+ T
T
kn k
ab1
k
k 1 = − ++
Proof
b
and
[ ( )]! ( )!!
( )! !!
( )! ( )!!
( )! !!
( )! !!
!( )! ( )!
( ) ( ) . . . ( ) ( ) . . . .( ) ( ) . . . ( ) ( ) . . .
T C a b
T C a b
T
T
C a b
C a b
n k kn a b
n k kn a
n k kn a
n k kn b
n k kn
nn k k
ab
n k n k k k kn k n k k k
ab
kn k
ab
1 1
1 11 1
1 3 2 1 1 2 3 2 11 3 2 1 1 2 3 2 1
1
( )
( )
( )
kn
kn k k
kn
kn k k
k
k
nk
n k k
nk
n k k
n k k
n k k
1
11 1
1
11 1
1 1
$ $ $ $ $
$ $ $ $
==
=
=
− − −
−
=
− + −
−
=−
− + −
=− − − − −
− + − − −
= − +
+−
−− − −
+
−− − −
−
− − −
−
b l
For this to work, k 0.≠
ExamplEs
1. Find the greatest coefficient in the expansion of .( )x3 10+
Solution
k10 −T C x
T C x
C x
3
3
3
( )k k
k
k kk k
kk k
110
101
10 1 1
101
11 1
===
+
−− − −
−− −
CONTINUED
´ ´
´
´
486 Maths In Focus Mathematics Extension 1 HSC Course
Comparing coefficients:
·
( ) ! !!
![ ( )] ! ( ) !
3(10 )(10 1) . . . 3 2 1 ( 1)( 2) . . . 2 1(10 1)(10 )(10 1) . . . 3 2 1 ( 1)( 2) . . . 2 1
T
T
C
C
C
C
CC
k kk k
k k k k kk k k k k
kk
kk
3
3
3
31
1010
3 1010 1 1
310 1
311
k
k
kk
kk
k
k
kk
1
101
11
10 10
101
10
1010
1
$ $ $ $
$ $ $ $
=
=
=
=−
− − −
=− − − − −
− + − − − − −
= − +
= −
+
−−
−
−
−
For the coefficient of
the coefficient of
i.e.
T T
T
T
kk
1
311 1
>
>
>
k k
k
k
1
1
−
+
+
,
.
k k
k
k
11 3
11 4
2 75
>>>
−
so for 1, 2,k = the coefficient of .T T>k k1+
For , , , , , ... ,k 3 4 5 6 7= the coefficient of .T T<k k1+
the term with the greatest coefficient occurs when 2k =
( )
T C x
T C x
T x
x
3
3
45 3
295 245
k kk k
110 10
2 110
210 2 2
38 2
2
====
+−
+−
so the greatest coefficient is 295 245.
2. Find the greatest coefficient in the expansion of .( )a3 2 7+
Solution
( )
( )
( )
T C a
C a
T C a
C a
C a
3 2
3 2
3 2
3 2
3 2
( )
k kk k
kk k k
k kk k
kk k
kk k k
17 7
7 7 7
71
7 1 1
71
8 1
71
8 1 8
=====
+−
− −
−− − −
−− −
−− − −
k 0.>
´
´´
\
487Chapter 10 The Binomial Theorem
Comparing coefficients:
C
C
C
C3 2
3 2
[ ( )] ! ( ) !!
( ) ! !!
( ) ! !( !)
![ ( )] ! ( ) !
( ) ! ![ ( )] ! ( ) !
3(7 )(7 ( 1)) . . . 3 2 1 ( 1)( 2) . . . 3 2 12(7 ( 1))(7 )(7 ( 1) . . . 3 2 1( 1)( 2)( 3) . . . 3 2 1
( ( ))
( )
T
T
k k
k k
k kk k
k kk k
k k k k kk k k k k k
kk
kk
kk
3
2
37 1 1
7
27
7
3 72 7
77 1 1
3 72 7 1 1
32 7 1
32 8
316 2
k
k
kk k
kk k
k
k
1
71
8 1
7 7
71
7
$ $ $ $ $
$ $ $ $
=
=
=
− − −
−
=−
− − −
=−
− − −
=− − + − −
− − − − + − − −
=− −
=−
= −
+
−− −
−
−
For the coefficient of T T>1k k+
the coefficient of 1T
T>1
k
k +
i.e. k
k
k k
k
k
316 2 1
16 2 3
16 5
351
>
>>
>
−
−
so for 1,k= 2 and 3, the coefficient of T T>1k k+ .For 4,k= 5, … the coefficient of T T<1k k+ .so the term with the greatest coefficient occurs when 3k= .this is the 4th term:
2
( )
( )
T C a
C a
C a
a
3 2
3
3 2
22 680
–4
73
7 3 3
73
4 3
73
4 3 4
4
====
so the greatest coefficient is 22 680.
k .0>
´
488 Maths In Focus Mathematics Extension 1 HSC Course
1. Find the 5th term of the
expansion of .x y2 3 27
+b l
2. Find the 4th term in the
expansion of .x x23
8
−b l
3. Find the coefficient of x2 in the
expansion of .( )xx
21841
r
rr
0
1818
=
−
ra ck m/
4. (a) Write out the expansion of
.xx3 3
5
+b l
(b) Find the greatest coefficient.
5. Find the term of x
x23
8
3−c m that is
independent of x.
6. Find the coefficient of x3 in each expansion.(a) ( )x 4 6−(b) ( )x3 8+
(c) )3C ( ) (x2 −kk
k k5
0
55
=
−/
(d) ( )x1 3 9−(e) ( )x 1 123 +
7. Find the constant term in each binomial product.
(a) xx23 62
+c m
(b) a
a10 13k
k k
0
10 10
2
3
=
−
ka c ek m o/
(c) ( )x6 2 7−
(d) x x4 12
−b l
(e) aa9 12k
k k
0
9 9 2
=
−
ka b ck l m/
8. Find the coefficient of a5 in the expansion of ( )a2 3 11− .
9. What is the term independent of
n in ( ) ?C n39
n21
kk
k
k9
0
9
2−
=c m/
10. Findthe coefficient of (a) y2 and the coefficient of (b) y5 in the
expansion of yy12
72
−f p .
11. Find the constant term of(a) ( )x4 8+(b) ( )x 6 3+
(c) x x8 2
r
rr
0
88
=
−
ra bk l/
(d) xx
3 2 9
2+c m
(e) yy12
10
3−e o
12. Find the coefficient of x3 in the expansion of
(a) C x 58
kk
k k
0
88
=
−/
(b) ( )x3 211r
r r
0
1111
=
−
ra k/
13. Find the coefficient of x in the
expansion of .x x3 7
−b l
14. Find the term independent of y
in the expansion of .yy
2 1 12
3+e o
15. the coefficients of x4 and x5 in the expansion of ( )x3 n− are equal in magnitude but opposite in sign. Find the value of n.
16. Find the values of a and b if the 5th term of ( )ax b 10+ is 13440x6 and the 8th term is -960x3.
17. in the expansion of ( ) ,ax b 6+ the coefficient of x4 is 2160 and the coefficient of x5 is 576− . Find a and b.
18. (a) simplify .C
C
k
k
81
8
−
(b) hence, or otherwise, find the greatest coefficient of the expansion of ( ) .x1 8+
10.4 exercises
489Chapter 10 The Binomial Theorem
19. (a) simplify C
C
k
k
61
6
−
.
(b) hence or otherwise, find the greatest coefficient in the expansion of ( )x2 6+ .
20. Find the greatest coefficient of(a) ( 3)x 9+(b) (2 1)a 10+(c) (4 3 )x 5+
(d) ( )x y28 k k
k
8
0
8−
= ka k/
(e) (3 1)n 11+(f) ( )a b2 6−(g) (3 4 )y 7+(h) ( )x y4 3 9−
(i) ( ) ( )t15 2 5k k
k
15
0
15−
= ka k/ (leave
your answer in coefficient and index form)(j) ( )n m4 7 12− (leave your answer in coefficient and index form)
21. Find the greatest coefficient of(a) ( )x 22 7+(b) ( )x2 1 5−(c) ( 3 )x y 9+
(d) x x22
8
+b l
(e) 2x x32
11
−b l
Further proofs
there is some further work you can do with coefficients and the expansion of the binomial product. some of these involve differentiation and integration.
Class Investigation
Find a pattern for the coefficient Cnk by using differentiation:
=
( )
( )
…
…
x C C x C x C x C x
x
C C C C C
C
10
1 0 0 0 0 0
1
Let
n n n n n nn
n
n n n n n nn
n
n
0 1 22
33
0 1 22
33
0`
+ = + + + + +=
+ + + + + +=
(1)
differentiating both sides of (1) gives
( )
( )
n x C C x C x n C xx
n C C C n C
n C
1 2 30
1 0 2 0 3 0 0
Let
n n n n nn
n
n n n n nn
n
n
11 2 3
2 1
11 2 3
2 1
1`
f
f
+ = + + + +=
+ = + + + +=
− −
− −
(2)
differentiating both sides of (2) gives
C C x
C C
C
0
+
+ +
( ) ( ) ( )
( ) ( ) ( )
( )( )
n n x n n C x
x
n n n n C
n n
n nC
1 1 2 3 2 10
1 1 0 2 3 2 1 0
1 2
21
Let
n n
nn
n nn
n
n
22 3
2
12 3
2
2
2
$
$
`
f
f
− + = + + −=
− + = + −− =−
=
− −
− −
n n
n n
n
Continue differentiating to find a general pattern for .Cnk
490 Maths In Focus Mathematics Extension 1 HSC Course
ExamplEs
1. Prove the sum of coefficients in ( )a b n+ is equal to .2n
Solution
( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
…
…
…
a b C a C a b C a b C b
a b
C C C C
C C C C
1
1 1 1 1 1 1 1 1
2
Let
n n n n n n n nn
n
n n n n n n n nn
n
n n n n nn
0 11
22 2
0 11
22 2
0 1 2`
+ = + + + += =
+ = + + + += + + + +
− −
− −
2. By comparing coefficients of x4 in both sides of (1 ) (1 ) (1 ) ,x x x8 4 4+ = + + show that
( )C Ckk
4 2 84
0
4
==/
Solution
Coefficient of x4 in ( )x1 8+ is .C84
terms of x4 in ( ) ( )x x1 14 4+ + are given byx· · · · · .C C x C x C C x C x C x C x C x C4
04
44 4
14
33 4
22 4
22 4
33 4
14
44 4
0+ + + +∴ coefficient in x4 is C C C C C C C C C C4
04
44
14
34
24
24
34
14
44
0+ + + +( ) ( ) ( ) ( ) ( )C C C C C4
02 4
12 4
22 4
32 4
42= + + + +
(since ,C C C C40
44
41
43= = )
( )Ck
k0
44 2=
=/
∴ )C C=(k
k4
0
42 8
4=/ (since coefficients of x4 are equal)
3. if ( )x xnr
1 n
r
nr
0+ =
=a k/
show that nr
3 4r
r
nn
0=
=a k/
Solution
.
.
( )
x
nr
nr
3
1 3 3
4 3
Let
Then
i.e.
n
r
nr
n r
r
n0
0
=
+ =
=
=
=
a
a
k
k
/
/
491Chapter 10 The Binomial Theorem
4. if n( ) ,x C x1 n
r
n
rr
0+ =
=/ show that .r C n2n
r
n
rn
1
1==
−/
Solution
( ) …x C C x C x C x C x1 n n n n n nn
n0 1 2
23
3+ = + + + + +differentiating both sides
( )
( ) ( ) ( ) ( )
n x C C x C x n C xx
n C C C n C
n C C C n C
r C
1 2 31
1 1 2 1 3 1 1
2 2 3
Let
n n n n nn
n
n n n n nn
n
n n n n nn
n
r
n
r
11 2 3
2 1
11 2 3
2 1
11 2 3
1
f
f
f
+ = + + + +=
+ = + + + += + + + +
=
− −
− −
−
=/
5. Given the expansion ( ) ,x x x xn n n nn
10 1 2
n n2 f+ = + + + +a a a ak k k k show
that … .( )
xn n x n x nn n
xnx
0 1 2 2 3 1 1
1 1n n2 3 1 1
+ + + + =+ +
+ −+ +
a a a ak k k k
Solution
integrating both sides:
( )
( )
x dx
nx
C
1
11
LHS n
n 1
1
= +
=+
++
+
#
x x x dx
x x xnx C
n n n nn
n n n nn2 3 1
0 1 2
1 2
RHS n
n
2
2 3 1
2
f
f
= + + + +
= + + + ++
++
0
a a a a
a a a a
k k k k
k k k k
#
n
n
1
1
+
+
( )
( )n
xC x x x
nx C
nx
C x x xnx
n n n nn
n n n nn
11
2 3 1
11
2 3 1
0 2
0 1 2
Son
n
1
2 3 1
2
3
2 3 1
f
f
++
+ = + + + ++
+
++
+ = + + + ++
+
+
1a a a a
a a a a
k k k k
k k k k
substitute :x 0=
( )n
Cn
nC
Cn
n n n nn1
1 00
20
30
10
11 0
11
0 1 2
n n1
3
2 3 1
3
3
f+
++ = + + + +
+
++ =
= −+
+ +
a a a ak k k k
( )
( )n
xn
x x n xnx
nx
x x xnx
n n nn
n n n nn
11
11
2 2 3 1
11 1
2 3 1
0 1
0 1 2
Son n
n n
1 2 3 1
1 2 3 1
f
f
++
−+
= + + + ++
++ −
= + + + ++
+ +
+ +
a a _ a
a a a a
k k i k
k k k k
492 Maths In Focus Mathematics Extension 1 HSC Course
ExamplE
Given that (1 ) ,x x x xn n n nn0 1 2
n n2 f+ = + + + +a a a ak k k k prove that n n n n
n0 1 22 2 2 3n n2 f+ + + + =a a a ak k k k .
Solution
( )x x x xn n n nn
10 1 2
n n2 f+ = + + + +a a a ak k k k
substitute :x 2=
+ +
23 + +
( ) ,n n n nn
n n n nn
1 2 2 2 20 1 2
0 1 2
n n
n n
2
2
f
f
+ = + +
2 2` = + +
a a a a
a a a a
k k k k
k k k k
1. Find the coefficient of x3 in the expansion of
(a) ( )31C xx
3kk
k
k
19 19
0
19
−−
=c m/
(b) ( ) ( )x x1 15+ +(c) ( ) ( )x x1 12 8+ +(d) ( ) ( )x x1 14 7+ +
2. By equating coefficients of ( )x1 6+ and ,( )x1 +( )x1 3 3+ show
that .26 3 3 3 3= +2 0 2 1 1a a a a ak k k k k
3. By equating coefficients of ( )x1 10+ and ( ) ( ) ,x x1 1 9+ + show
that .10 9 9= +4 4 3a a ak k k
4. By equating coefficients of ,xk use ( ) ( )x x1 1= + +( )x1 n n1+ + to prove that .C C Cn
kn
kn
k1
1= ++−
5. (a) Factorise .( )x1 6+ +( )x1 5+Find the coefficient (b)
of x3 in the expansion of .( )x1+ +( )x1 5 6+
6. Find the coefficient of x2 in
( )x xx 11 35+ +b l .
7. the coefficients of x4 and x5 are equal in the expansion of
.( )x1 n+ evaluate n.
10.5 exercises
in order to do questions like these examples, it is a good idea to look at what you are trying to prove first.
Notice that in example 4, the result to be proven included an .x –n 1
this means that we use differentiation since .( )x nxdxd n n 1= − also, when
differentiating, the constant term is zero, so the sum of terms starts from k 1= rather than .k 0=
in example 5, the result included .xn 1+ this means that we use integration
since .x dx Cnx
1n
n 1
= ++
+
# remember with integration that we need to evaluate C,
so we use a substitution for x, usually ,x 0= to evaluate it.
You can see from the examples, that we quite often use substitution of different values of x in these types of questions. sometimes this is all that is needed.
493Chapter 10 The Binomial Theorem
8. Given that n( ) ,x xnk
1 k
k
n
0+ =
=a k/
show that
nk
6 7k
k
nn
0=
=a k/(a)
(b) nk
2 3k
k
nn
0=
=a k/
nk
3 2k
k
nn
0
2==a k/(c)
9. if n( ) ,x xnk
1 k
k
n
0+ =
=a k/ use
differentiation to show that
.k nk
n2k
nn
1
1==
−a k/
10. Given ( ) ,a x a xnk
n n k k
k
n
0+ = −
=a k/
show that
(a) n 1=0a k
(b) n( )a an1 n k
k
n
0+ = −
= ka k/
(c) ) a1( ) (a n1 n k n k
k
n
0− = − −
= ka k/
(d) ( )n a x k a xnn
k
nn k k1
1
1+ =−
=
− −
ka k/
(e) nn =1a k
11. if n( )x1 +
…x x xn n n nn0 1 2
n2= + + + +a a a ak k k k
show that … ( )n n n n n
n2 4 8 2
0 1 2 3n− + − + + −a a a a ak k k k k
( ) .1 n= −
12. Given that ( )x1 n+
… ,x x xn n n nn0 1 2
n2= + + + +a a a ak k k k
show that
(a) …n n n nn
20 1 2
n+ + + + =a a a ak k k k
(b) … ( )n n n nn
1 00 1 2
n− + − + − =a a a ak k k k
13. Given ( ) ,x xn1 2k
nk2
0
2
+ ==
n
ka k/
(a) n 42showk
nn
0
2
== ka k/
(b) by differentiating both sides,
k nn 42show thatk
nn
1
2
== ka k/
14. if ( ) ,x xnr
1 n
r
nr
0+ =
=a k/
show by integration that
r( )
.r
nr n1
1
11
r
n
0=
+−
+=a k/
15. show that
…n n nn
nn0 2
11 3
12 1
1+ + + ++a a a ak k k k
n 1
2 1n 1
= −+
+ by integrating ( )x1 n+ .
494 Maths In Focus Mathematics Extension 1 HSC Course
1. Write the binomial expansion of x x3 12
+b l in sigma notation.
2. if ( ) ,x xnr
1r
nr n
0= +
=a k/ show that
.r nnr
2r
nn
1
1==
−a k/3. Find the coefficient of x5 in the
expansion of .( )x3 2 7−
4. By equating coefficients of ( )x1 8+ and ( ) ( ) ,x x1 13 5+ + show that
.83
33
50
32
51
31
52
30
53
= + + +b b b b b b b b bl l l l l l l l l
5. if n( ) ,x x x xn n n nn
10 1 2
n2 f+ = + + + +a a a ak k k k show that
(a) n 10
=a k
(b) )1 0=n(n n n n nn0 1 2 3
f− + − + −a a a a ak k k k k
6. Find the constant term of
.x x12 1
k
kk
0
1212
=
−kb al k/
7. simplify .( ) ( )3 1 3 15 5+ − −
8. use ( ) ( ) ( )x x x1 1 1n n 1+ = + + − to prove
that kn n1 1− − .n
k = + k 1−a a ak k k
9. expand .( )x y2 3 5+
10. Find the coefficient of x3 in the expansion of ( ) ( ) .x x1 14 5+ +
11. Find the term independent of x in
.( )xx 314 2 kk
0
14 14 −
kb al k/
12. Find the greatest coefficient of
.x210
r
r r
0
1010
=
−rb l/
13. if the coefficient of x3 is 4320 and the coefficient of x4 is −1620 in the expansion of ( ) ,ax b 5+ find a and b.
14. (a) simplify .C
C
k
k
101
10
−
hence, or otherwise, find the greatest (b) coefficient of .( )x3 10+
15. the coefficients of x2 and x3 in the expansion of ( )x2 n− are equal in magnitude but opposite in sign. evaluate n.
16. Find the coefficient of x4 in the expansion of .8( ) ( )x x1 1 7+ + +
17. Find the greatest coefficient in the expansion of .( )x3 2 9+
18. expand
(a) ( )pn 3 2n k
k
k
0
5−
= ka k/
(b) ( )5 2 6−
(c) 3xx2
4
−c m
(d) )C b( )a2 n k−n (kk
k
0
3
−=/
(e) ( )2 3 2 7−
19. if ( ) a b5 3 5 35− = + , evaluate a and b.
20. Given ( ) ,a x a xnk
n n k k
k
n
0+ = −
=a k/ show that
(a) ( )a ank
2 2n k n k
k
n
0+ = −
=a k/
(b) ( )n a k ank
1 –n
k
nn k1
1+ =
=
−a k/
(c) ( )
na x
na x an
k 1 1
n k k
k
n n n1
0
1 1
+=
++ −− +
=
+ +
a k/
Test Yourself 10
495Chapter 10 The Binomial Theorem
Challenge Exercise 10
1. By considering coefficients of x4 on both sides of ( ) ( ) ( ) ,x x x1 1 120 10 10+ = + + show that
.( ) ( )· ·C C C C C C2204
100
104
101
103
102
2= + +
2. (a) show that
.n2( ) 1 ( )x x xx1 11n nn
+ + = +a k
show that (b)
… .n n n nn
nn0 1 2
22 2 2 2
+ + + + =a a a a bk k k k l
3. By equating coefficients on both sides of ( ) ( ) ( ) ,x x x1 1 1n n n2+ − = − show
)1 0=2
( nkk
nk
0−
=a k/ if n is odd.
4. if ( )x C x3 2 kk
k15
0
15
+ ==/
show (a) ( )( )
C
C
kk
3 12 15
k
k 1 =+−+
find the greatest coefficient (b) Ck (leave your answer in index form)
5. show that k
.k n12
2 13 1k n
k
n 1 2 1
0
2
=+ +
−+ +
=
n2c m/
6. Prove that 1nr r n1
21
3
r
n r n
0
1 1
=+ +
−=
+ +a k/ by
finding the integrals
n( )x dx10
2+# and .x dxn
rr
nr
00
2=
=a k/#
7. Prove C· · …( )( )
( ) ( )…( )k k
n n n n k1 2 3 1
1 2 1n
k =−
− − − +
using mathematical induction.