Introduction To Binomial Theorem  arbindsingh.com · Introduction To Binomial Theorem A Binomial...
Embed Size (px)
Transcript of Introduction To Binomial Theorem  arbindsingh.com · Introduction To Binomial Theorem A Binomial...

Introduction To Binomial Theorem
A Binomial Expression
Any algebraic expression consisting of only two terms is known as a binomial expression. It's expansion in power of x is shown as the binomial expansion.
For example: (i) a + x (ii) a2 + 1/x2 (iii) 4x  6y
Binomial Theorem
Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.
It can be easily understood by examples.
(a + x)2 = a2 + 2ax + x2
(a + x)2 = a3 + 3a2x + 3ax2 + x3
Here, we see that the expression of (a + x)2 is simple, we just multiply (a + x) by (a + x). Expansion of (a + x)3 is little tougher, but what happens when the expansion is raised to the power of ten or more? So, we have to establish the formula for (a + x)n, where n is any integer. Let us define 'a' as the first term, 'x' as the second term and 'n' as the exponent.
The total terms in the expansion of (a + x)2 and (a + x)3 are 3 and 4 respectively, which means that the number of terms in the expansion is one more than the exponent. So total number of terms in the expansion (a + x)n is (n + 1).
Now, for n = 2
(a + x)2 = a2x0 + 2a1x1/1 + (2(21)/1*2)a0x2
= (F.T.)n (S.T.)0 + n/1! (F.T.)(n1) (S.T.)1 + (n(n1)/2!) (F.T.)(n2) (S.T.)2
= a2 + 2ax + x2
Note: F.T. refers to first term i.e. 'a' and S.T. refers to second term i.e. 'x'
Similarly,
(a + x)3 = a3 + 3a2x + ((3(31)/(1*2))a0 x2 + ((3(31)(32))/(1*2*3))x3
= a3 + 3a2x + 3ax2 + x3
When n is a positive integer, then
ArbindText Boxwww.arbindsingh.com imparting education to the world....

(a + x)n = nC0an + nC1an1x + nC2an2 x2 +...+ nCranr xr +...+ nCnxn,
Where nC0 . nC1 . nC2 ... nCn are called Binomial coefficients.
Introduction To Binomial Theorem
Proof of Binomial Theorem
Proof of Binomial Theorem is very simple; we can prove it by using the mathematical induction.
Proof:
Step I: Let n = 1
L.H.S. = a + x
R.H.S. = a + 1C1 x = a + x
So, theorem is true for n = 1.
Step II:
Let the theorem be true for n = m, than
P(m) : (a + x)m = mC0 am + mC1 am1 x1 + mC2 am2 +...+ mCm xm (i)
Step III:
We have to prove for n = m + 1 i.e. we have to prove that
P(m + 1): (a + x)m+1 = m+1C0 am+1 + m+1C1 am x1
+...+ m+1Cm a xm + m+1Cm+1 xm+1 ....(ii)
Multiplying by (a + x) on both sides in equation (i), we get,
(a + x)m+1 = (mC0 am+1 + mC1 am +...+ mCm1 a2 xm1 + mCm a xm)
= (mC0 am x + mC1 mC0) am x +...+ mCm1 a xm + mCm xm+1)
Or, (a+x)m+1 = mC0 am+1 + (mC1 + mC0) am x +...+ (mCm1 + mCm2) a2 xm1

+ (mCm + mCm1) a xm + mCm xm+1
= m+1C0 am+1 + m+1C1 am x + m+1C2 am1 x2 +...+
m+1Cm1 a2 xm1 + m+1Cm a xm + m+1Cm+1 xm+1.
Hence, Proved.
Illustration:
Expand (x + 1/x)7 .
Solution:
(x + 1/x)7 = 7C0x7 + 7C1x6 (1/x) + 7C2x5 (1/x2) + 7C3x4 + 1/x3 + 7C4x4
(1/x3) + 7C5x2 + (1/x5) + 7C6x (1/x6) + 7C7 (1/x7)
= x7 + 7x5 + 21x3 + (35 x) + (35/x )+( 21/x3) + (7/x5)+ (1/x7) .
Properties Of Binomial Expansion
1. There are (n + 1) terms in the expansion of (a + b)n, the first and the last term being an and bn respectively. If nCx = nCy, then either x = y or x + y = n
=> nCr = nCnr = n!/r!(nr)! .
2. The general term in the expansion of (a + x)n is (r + 1)th term given as Tr+1 =nCranr + xr.
Similarly the general term in the expansion of (x + a)n is given as Tr+1 = nCrxnr ar. The terms are considered from the beginning.
3. The binomial coefficient in the expansion of (a + x)n which are equidistant from the beginning and the end are equal i.e. nCr = nCnr.

Note: Here we are using nCr + nCr1 = n+1Cr this concept will be discussed later in this chapter. Also, we have replace mC0 by m+1C0 because numerical value of both is same i.e. 1. Similarly we replace mCm by m+1Cm+1.
Illustration:
Find the expansion of (ax)n.
Solution: We know that
(a + x)n = an + na(n1) x + (n(n1)/2!) a(n2) x2 +...+ xn.
putting x = x in the above expansion, we get,
(ax)n = [a + (x)]n = an + na(n1) (x) + (n(n1)/2!) a(n2) (x)2 +......+ (x)n.
= an  na(n1) x + (n(n1)/2!) a(n2) x2 ......+ (1)n xn.
Illustration:
Find the value of (a+√(a2 1))7 +(a√(a2 1))7.
Solution:
Here, we have to find the sum of two expansions whose terms are numerically the same, but in the second expansion the second, fourth, sixth and eight terms are negative, and therefore cancel the corresponding terms of the first expansion. Hence, the given expression
= 2 {a7 + 21 a5 (a2  1)+ 35 a3(a2  1)2 + 7a (a2  1)3}
= 2a (64 a6  112 a4 + 56 a2  7)

Binomial Coefficients
We know that,
(a + x)n = nC0 an + nC1 an1 x + nC2 an2 x2 +...+ nCn xn ... (i)
Let us write equation (i) in particular form by putting a = 1. Now it becomes,
(1 + x)n = nC0 + nC1 x + nC2 x2 +...+ nCr xr +...+ nCn xn ... (ii)
Coefficients attached with different powers of x are called Binomial Coefficients.
Now, again put a = 1 in the expression
(a + x)n = an + n an1 x + (n(n1))/2! an2 x2 + (n(n1)(n2))/3! an3 x3
+...+ (n(n1)(n2)....(nr+1))/r! anr xr +......+ xn
we get,
(1+x)n = 1 + nx + (n(n1))/2! x2 +...+ (n(n1)(n2)......(nr+1))/r! xr +...+ xn ...(iii)
If we compare coefficient of xr in expansions (ii) and (iii), we have,
nCr = (n(n1)(n2)......(nr+1))/r! = n!/r!(nr)!
What does it mean when we compare two expansions?
Since expansion (ii) is valid for any value of x, we can replace x by 1/x , (x ≠ 0) in

it, we get,
(1 + (1/x))n = nC0 + nC1 1/x + nC2 1/x2 +...+ nCr 1/xr +...+ nCn 1/xn
Multiplying both sides by xn, it becomes,
(1+x)n = nC0 xn + nC1 xn1 +...+ nCr xnr +...+ nCn ... (iv)
Compare expansion (ii) and (iv). It will give us, that
nCr = nCnr, this is a beautiful result, which says that in the expansion of the equations of the like of (ii) the rth coefficient from the beginning is equal to the rthcoefficient from the end.
Sum Of Binomial Coefficients
Putting x = 1 in the expansion (1+x)n = nC0 + nC1 x + nC2 x2 +...+ nCx xn, we get,
2n = nC0 + nC1 x + nC2 +...+ nCn.
We kept x = 1, and got the desired result i.e. ∑nr=0 Cr = 2n.
Note: This one is very simple illustration of how we put some value of x and get the solution of the problem. It is very important how judiciously you exploit this property of binomial expansion.
Illustration:
Find the value of C0 + C2 + C4 +... in the expansion of (1+x)n.
Solution:
We have,
(1 + x)n = nC0 + nC1 x + nC2 x2 +... nCn xn.
Now put x = x; (1x)n = nC0  nC1 x + nC2 x2 ...+ (1)n nCn xn.
Now, adding both expansions, we get,

(1 + x)n + (1x)n = 2[nC0 + nC2 x2 + nC4 x4 +.........]
Put x = 1
=> (2n+0)/2 = C0 + C2 + C4 +......
or C0 + C2 + C4 +......= 2n1
We have found the sum of binomial coefficients. But if these coefficients are multiplied by some factors can we find the sum for such expressions?
Yes, we can often find it by creatively applying what we have learnt. Let us see how differentiation and integration are useful in these situations.
Suppose we want to calculate the value of
∑nr=0 Cr i.e. 0.C0 + 1 C1 + 2C2 +...+ n nCn
If we take a close look to the sum to be found, we find that coefficients are multiplied with respective powers of x.
If we want to multiply the coefficient of x by its power differentiation is of help. Hence differentiate both sides of
(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 +...+ nCn xn, with respect to x we get
n(1 + x)n1 = 1 C1 x11 + 2 C2 x21 +...+ n Cn xn1
Put x = 1, we get, n 2n1 = + 1 C1 + 2 C2 +...+ n Cn.
Or, ∑nr=0 r Cr = n 2n1, which is the answer.
Now, think how to find the following sum
C0/1 + C1/2 + C2/3 +......+ Cn/n+1 = ?
In this sum coefficients are divided by the respective power of x + 1. This expression can be achieved by Integrating the expansion of (1 + x)n under proper limits.

Important: It has said earlier in this chapter that we should use and exploit the property that x can take any value in the expansion of (1 + x)n.
Now, let us try to find the value of
C0  C2 + C4C6 +.........
Analysing the above, expression, we find that C0, C2, C4 are all coefficients of even powers of x.
Had it been like C0 + C2 + C4 +...... we could have evaluated simply by the method described earlier or had it been like C0C1 + C2C3 +... we could have put x = 1 in the expansion of (1 + x)n and find the sum.
But here the case is different. The expression consists of coefficients of only even powers. In such cases when there is alternative sign charge for the coefficient, of same nature of powers (even or odd) use of I (iota) comes to our rescue.
(1 + x)n = C0 + C1 x + C2 x2 +......
(1  x)n = C0  C1 x + C2 x2 +......
By adding (1 + x)n + (1  x)n = 2[C0 + C2 x2 + C4 x4 +......].
Clearly use of √1 (iota) will generate the given expression.
((1+i)n + (1i)n)/2 ] [C0  C2 + C4 + ......]
Coefficient Of A Particular Term

In the expression of (a + x)n, the coefficient of second term is nC1, of the third term is nC2, of he fourth term is nC3 and so on. The suffix in each term being one less than the number of the term to which it applies; hence nCr is the coefficient of the (r+1)th term. This is called the General Term, because by giving different numerical values to r any of the coefficients may be found. Also, the indices of a and x in the (r + 1)th term are expressable in terms of r.
Thus, General term of the expansion (a + x)n is dented as
Tr+1 = nCr anr xr = ((n(n1)(n2)....(nr+1))/r!) anr xr.
Note :General term is useful in finding the following in Binomial expression. (a) Particular term. (b) Middle term (c) Term independent of x. (d) Term containing the greatest coefficient of x.
Illustration:
Find the fifth term in the expansion of (a + 2x3)17.
Solution:
General term of the expansion (a + x)n is
Tr+1 = nCr anr xr.
Now, for the given expression, required term is
T5 = 17C4 a13 (2x3)4.
= 38080 a13 x12.
Illustration:
Find the term independent of x in (1 + x + 2x3) ((3/2)x2  (1/3x))9.
Solution:
We want to find the term independent of x in product of (1 + x + 2x3) and

((3/2)x2  (1/3x))9 . So, we should find the terms containing x0, x1 and x3 in the expansion ((3/2)x2  (1/3x))9 and multiply them respectively with the appropriate term from (1 + x + 2x3) i.e. constant term i.e. 1, term containing x i.e. 1 and the term containing x3 i.e. 2.
Now, Tr+1 = nCr anr xr [this is general term in (a+x)n]
= 9Cr ((3/2)x2)9r ((1/3x))r
This is general term in [((3/2)x2  (1/3x))9]
(i) to find the term containing x0 we use 18  3r = 0, i.e. r = 6
so coefficient of x0 we use (1)6 9C6 33 23 = 9C6 (1/22.33).
(ii) to find the term containing x1 we use 183r = 1 i.e. r = 19/3 ( I) which is impossible.
(iii) to find the term containing x3 we put 183r = 3 i.e. r = 7. So, coefficient of x3 in ((3/2)x2  (1/3x))9 = (1)7 9C7 35 22 = 9C7 (1/23.35) .
Therefore, required coefficient = 9C6 (1/23.33)× 12 × 9C7 (1/23.35) = 17/54.
Illustration:
Find the coefficient of x24 in (x2 +3a/x)15.
Solution:
The general term ((r + 1)th term) is
(x2 +3a/x)15 = 15Cr(x2)15r (3a/x)r = 15Cr x302r (3rar/xr) = 15Cr 3r ar x303r
If this term contains x24, then 303r = 24
=> 3r = 6 => r = 2.
Therefore, the coefficient of x24 = 15C2 9a2.
Illustration:
If the binomial coefficient of the (2r + 4)th term and the (r  2)th term in the expansion of (1 + x)18 are equal, find the value r.
Solution:
The coefficient of (2r + 4)th term in (1 + x)18 = 18C2r+3 and the coefficient of (r

2)th term = 18Cr3, so that 18C2r+3 = 18Cr3.
Either 2r + 3 = r  3, => r = 6,
Or 2r + 3 + r  3 = 18 => 3r = 18 => r = 6. (Since r ε N)
Illustration:
If the 4th term in the expansion (px + 1/x)n is independent of x, find the value of n. Also calculate p if the 4th term is 5/2.
Solution:
Here T4 = T3+1 = nC3(px)3 (1/x)n3 = nC3 p3 x6n.
T4 is independent of x => 6  n = 0 or n = 6.
Now given T4 = 5/2 => 6C3. p3 = 5/2 => p3 = 5/2 . = 1/8 => p = 1/2.
Illustration:
If the expansion of (1 + x)43 the coefficient of (2r + 1)th term is equal to the coefficient of (r + 2)th term find r.
Solution:
Given in the expansion of (1 + x)43 the coefficient of (2r + 1)th term = the coefficient of (r + 1)th term.
43C2r = 43Cr+1. Either 2r = r + 1 => r = 1,
or 2r + r + 1 = 43 => r = 14.
Hence r =1, 14
Illustration:
Find the coefficient of x3 in the expansion of (1 + x + 2x2) (2x2  (1/3x))9.
Solution:
Tr+1 in (2x2  (1/3x))9 is 9Cr (2x2)9r (1/3x)r = 9Cr29r (1/3)r x183r
=> coefficient of x3 is 9C5.24 (1/3)5 = 224/7.

Middle Term
(i) When n is even
Middle term of the expansion is the (n/2 + 1)th term
i.e. nCn/2 an/2 in the expansion of (a + b)n.
(ii) When n is odd
Middle terms of the expansion are the (n/2 + 1)th term and the ((n+3)/2)th term.
These are given by,nC((n1)/2) a((n+1)/2) a((n1)/2) and nC((n+1)/2) a((n1)/2)a((n+1)/2) in the expansion of (a + b)n.
e.g. middle term in the expansion of (1+x)4 and (1+x)5.
Expansion of (1+x)4 have 5 terms, so third term is the middle term which is the ((4/2)+1)th term.
Expansion of (1+x)5 have 6 terms, so 3rd and 4th both are the middle terms, which are the ((5+1)/2)th and ((5+3)/2)th terms.
Note: * rth term from the end = (n  r + 2)th term from the beginning. * If there are two middle terms, then the binomial coefficients of two middle terms will be equal and those two coefficients will be greatest.
Illustration:
Find the middle term in the expansion of (1  2x + x2)n.
Solution:
We have (1  2x + x2)n = [(1 x)2]n = (1  x)2n.
Here 2n is an even integer =>((2n/2) + 1)th i.e. (n+1)th term will be the middle term.
Now (n+1)th term in (1x)2n = 2nCn (1)2nn(x)n = 2nCn(x)n = (2n)!/n!n! (x)n.
Greatest Binomial Coefficient

To determine the greatest coefficient in the binomial expansion, (1+x)n, when n is a positive integer. Coefficient of
(Tr+1/Tr) = Cr/Cr1 = (nr+1)/r = ((n+1)/r)  1.
Now the (r+1)th binomial coefficient will be greater than the rth binomial coefficient when, Tr+1 > Tr
=> ((n+1)/r)1 > 1 => (n+1)/2 >r. ....... (1)
But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given by the greatest value of r, consistent with (1) i.e., r = n/2 and hence the greatest binomial coefficient is nCn/2.
Similarly in n be odd, the greatest binomial coefficient is given when,
r = (n1)/2 or (n+1)/2 and the coefficient itself will be nC(n+1)/2 or nC(n1)/2, both being are equal
Note: The greatest binomial coefficient is the binomial coefficient of the middle term.
Illustration:
Show that the greatest the coefficient in the expansion of (x + 1/x)2n is (1.3.5...(2n1).2n)/n! .
Solution:
Since middle term has the greatest coefficient.
So, greatest coefficient = coefficient of middle term
= 2nCn = (1.2.3...2n)/n!n! = (1.3.5...(2n1).2n)/n!.
Numerically greatest term
To determine the numerically greatest term in the expansion of (a + x)n, where n is a positive integer.

Consider
Thus
Note : {((n+1)/r)  1} must be positive since n > r. Thus Tr+1 will be the greatest term if, r has the greatest value as per the equation (1).
Illustration:
Find the greatest term in the expansion of (32x)9 when x = 1.
Solution:
Tr+1/Tr = ((9r+1)/r) . (2x/3) >1
i.e. 20 > 5r
If r = 4, then Tr+1 = Tr and these are the greatest terms. Thus 4th and 5thterms are numerically equal and greater than any other term and their value is equal 39 × 9C3 × (2/3)3 = 489888.
Illustration:
Find the greatest term in the expansion of (2 + 3x)9 if x = 3/2.
Solution:
Here Tr+1/Tr = ((nr+1)/r)(3x/2) = ((10r)/r)(3x/2), (where x = 3/2)
= ((10r)/r)(3x/2)(3/2) = ((10r)/r).9/4 = (909r)/4r
Therefore Tr+1 > Tr, if 90  9r > 4r.

=> 90 > 13r => r < 90/13 and r being an integer, r = 6.
Hence Tr+1 = T7 = T6+1 = 9C6 (2)3 (3x)6 = 313.7/2.
Illustration:
Given that the 4th term in the expansion of (2 + (3/8)x)10 has the maximum numerical value, find the range of values of x for which this will be true.
Solution:
Given 4th term in (2 + (3/8)x)10 = 210 (1 + (3/16)x)10, is numerically greatest
=>T4/T3 > 1 and T5/T4 < 1
=>
=>x > 2 and x < 64/21
=> x ε [(64/21),2]U[2,(64/21)].
Illustration:
Find the greatest term in the expansion of √3(1 + (1/√3))20.
Solution:
Let rth term be the greatest
Since Tr/Tr+1 = (r/(21r)).√3.
Now Tr/Tr+1 > 1 => r > 21/(√3+1) ...... (1)
Now again Tr/Tr+1 = (r/(21r)) √3 < 1 => r < (22+√3)/(√3+1) ...... (2)
from (1) and (2)
22/(√3+1) < r < (22+√3)/(√3+1)
=> r = 8 is the greatest term and its value is √3 . 20C7 (1/√3)7 = 20C7 (1/27).

Particular Cases
We have (a + x)n = an + nC1 an1 x + nC2 an2 xr +...+ xn. ...... (1)
(i) Putting x = x in (1), we get
(ax)n = an  nC1 an1x + nC2 an2 x2  nC2 an3 x3 +...+ (1)r nCr anr xr+...+ (1)n xn.
(ii) Putting a = 1 in (1), we get,
(1+x)n = nC0 + nC1x + nC2x2 +...+ nCr xr +...+ nCnxn. ... (A)
(iii) Putting a = 1, x = x in (1), we get
(1x)n=nC0nC1 x + nC2 x2nC3x3 +... (1)r nCr xr +... (1)n nCnxn ... (B)
Tips to Remember
(a) Tr/Tr+1 = ((nr+1)/r ).(x/a) for the binomial expansion of (a + x)n.
(b) (n+1)Cr = nCr + nCr1.
(c) r nCr = n n1Cr1
(d)
(e) When n is even,
(x + a)n + (x  a)n = 2(xn + nC2 xn2 a2 + nC4 xn4 a4 +...+ nCn an).
When n is odd,
(x + a)n + (x  a)n = 2(xn + nC2 xn2 a2 +...+ nCn1 x an1).
When n is even
(x + a)n  (x  a)n = 2(nC1 xn1 a + nC3 xn3 a3 +...+ nCn1 x an1).
When n is odd
(x + a)n  (x  a)n = 2(nC1 xn1 a + nC3 xn3 a3 +...+ nCn an).
Properties of Binomial Coefficients

For the sake of convenience, the coefficients nCo, nC1, ..., nCr, ..., nCn are usually denoted by Co, C1,..., Cr, ..., Cn respectively
Put x = 1 in (A) and get, 2n = Co + C1 + ... + Cn ... (D)
Also putting x = 1 in (A) we get,
0 = Co  C1 + C2  C3 + ......
=> C0 + C2 + C4 + ...... + C1 + C3 + C5 +...... = 2n.
Hence Co + C2 + C4 +...... = C1 + C3 + C5 +...... = 2n1.
Illustration:
If (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, then prove that C0 + (Co + C1) + (C0 + C1+ C2) + ... (C0 + C1 + C2 +...+ Cn1) = n2n1 (where n is an even integer.
Solution:
C0 + (C0 + C1) + (C0 + C1 + C2) +... (C0 + C1 + C2 +...+ Cn1)
= C0+(C0 + C1 + C2 +...+ Cn1)+(C0 + C1)+(C0 + C1 + C2 +...+ Cn2)+...
= (C0 + C1 + C2 +...+ Cn)+ (C0 + C1 + C2 +...+ Cn)+... n/2 times
= (n/2)2n = n . 2n1
Some Important Results
1. Differentiating (1+x)n = C0 + C1x + C2x2 +...+ Cnxn of both sides we have,
n(1 + x)n1 = C1 + 2C2x + 3C3x2 +...+ nCnxn1. ... (E)
Put x = 1 in (E) so that n2n+1 = C1 + 2C2 + 3C3 + ...+ nCn.
Put x = 1 in (E) so that 0 = C1  2C2 +...+ (1)n1 nCn.
Differentiating (E) again and again we will have different results.
2. Integrating (1 + x)n, we have,
((1+x)n+1)/(n+1) + C = C0x + C1x2/2 + C2x3/3 +.....+Cnxn+1/(n+1) (where C is a constant)

For x = 0, we get C = 1/(n+1) .
Therefore ((1+x)n+1  1)/(n+1) = C0x + C1x2/2 + C2x3/3 +.....+Cnxn+1/(n+1) ..... (F)
Put x = 1 in (F) and get
(2n+1  1)/n+1= C0 + C1/2 +...Cn/n+1 .
Put x = 1 in (F) and get, 1/n+1 = C0  C1/2 + C2/3  ......
Put x = 2 in (F) and get, (3n+11)/n+1 = 2 C0 + 22/2 C1 + 22/3 C2 +...+ 2n+1/n+1 = Cn.
Problems Related to Series of Binomial Coefficients in Which Each Term is a Product of an Integer and a Binomial Coefficient, i.e. In the Form k.nCr.
Illustration:
If (1+x)n = xr then prove that C1 + 2C2 + 3C3 +...+ nCn = n2n1.
Solution:
Method (i) : rth term of the given series
rth term of the given series, tr = nCr
=> tr = r × n/r × n1Cr1 = n × n1Cr1 (because nCr =n/r .n1Cr1)
Sum of the series =
Put x = 1 in the expansion of (1 + x)n1, so that
(n1C0 + n1C1 +...+ n1Cn1) = 2n1
=> = n.2n1.
Method (ii) : By Calculus
We have (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn. ... (1)
Differentiating (1) w.r.t. x, we get

n(1 + x)n1 = C1 + 2C2x + 3C3 x2 +...+ n Cnxn1. ... (2)
Putting x = 1 in (2), we have, n 2n1 = C1 + 2C2 +....+ nCn. ... (3)
Illustration:
If (1+ x)n = xr then prove that C0 + 2.C1 + 3.C2+...+(n+1)Cn=2n1(n+2).
Solution:
Method (i): rth term of the given series
rth term of the given series
tr = nCr1 = [(r1) + 1]. nCr1
= (r1) nCr1 + nCr1 = n. n1Cr2 + nCr1 (because nCr1 =n/(r1) . n1Cr2)
Sum of the series =
= n[n1C0 + n1C1 +...+ n1Cn1]+[nC0 + nC1 +...+ nCn] = n.2n1 + 2n = 2n1 (n+2).
Method (ii) by Calculus.
We have (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn. ... (1)
Multiplying (1) with x, we get
x(1+x)n = C0x + C1x2 + C2x3 +...+ Cnxn+1. ... (2)
Differentiating (2) w.r.t. x, we have
(1 + x)n + n(1 + x)n1 x = C0x + 2C1x2 +...+ (n+1)Cnxn ... (3)
Putting x = 1 in (3), we get
2n + n.2n1 = C0 + 2C1 + 3C2 +...+ (n+1)Cn
=> C0 + 2C1 + 3C2 +...+ (n+1)Cn = 2n1 (n+2).

Problems Related to Series of Binomial Coefficient in Which Each Term is Binomial Coefficient divided by an Integer, i.e. in the Form of nCr/k.
Illustration:
If (1+x)n =
Solution:
Method (i) : rth term of the given series
Method (ii): By Calculus
(1+x)n = C0 + C1x + C2x2 +...+ Cnxn ... (1)
Integrating both the sides of (1) w.r.t. x between the limits 0 to x, we get
... (2)
Substituting x = 1 in (2), we get 2n+1/n+1 = C0 + c1/2 + c2/3 +.....+ cn/n+1 .
Illustration:
If (1+x)n = xr ,show that
Method I : rth term of the given series Tr =

Method II : (By Calculus)
(1 + x)n = C0 + C1x + C2x2 +...+ Cnxn
=> x(1+x)n = C0x + C1x2 + C2x3 +...+ Cnxn+1 ... (1)
Integrating both the sides of (1) with respect to x
Put x = 0, => k = 1/((n+1)(n+2))
Put x = 1,
.
Problem Related to Series of Binomial Coefficients in Which Each Term is a Product of two Binomial Coefficients.
(a) If sum of lower suffices of binomial expansion in each term is the same
i.e. nC0 nCn + nC1 nCn1 + nCn2 +...+ nCn nC0
i.e. 0 + n = 1 + (n1) = 2 + (n2) = n + 0.
Then the series represents the coefficients of xn in the multiplication of the following two series

(1+x)n = C0 + C1x + C2x2 +...+ Cnxn
and (1+x)n = C0 + C1x + C2x2 +...+ Cnxn.
Illustration:
Prove that C0Cr + C1Cr+1 + C2Cr+2 +...+ Cnr Cn = (2n)!/((nr)!(n+r)!) .
Solution:
We have,
C0 + C1x + C2x2 + ... + Cnxn = (1+x)n ... (1)
Also C0xn + C2xn2 +...+ Cn = (x+1)n ... (2)
Multiplying (1) and (2), we get
(C0 + C1x2 +...+ Cnxn)(C0xn + C1xn2 + C2xn2 +...+ Cn) = (1+x)2n ... (3)
Equating coefficient of xnr from both sides of (3), we get
C0Cr + C1Cr+1 + C2Cr+2 +...+ CnrCn = 2nCnr = (2n)/((nr)!(n+r)!) .
Illustration:
Prove that Co2 + C12 +...+ Cn2 = (2n)!/n!n! .
Solution:
Since
(1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, ... (1)
(x + 1)n = C0xn + C1xn2 +...+ Cn, ... (2)
(C0 + C1x + C2x2 +...+ Cnxn)(C0xn + C1xn1 + C2xn2 +...+ Cn)=(1+x)2n.
Equating coefficient of xn, we get
C02 + C12 + C22 +...+ Cn2 = 2nCn = (2n)!/n!n! .

Illustration:
If (1+x)n = , then prove that
mCr nC0 + mCr1 nC1 + mCr2 nC2 +...+ mC1 nCr1 + mC0 nCr = m+nCr
where m, n, r are positive integers and r < m and r < n.
Solution:
(1+x)n = nC0 + nC1x + nC2x2 +...+ nCrxr +...+ nCnxn ... (1)
and also
(1+x)m = mC0 + mC1x + mC2x2 +...+ mCrxr +...+ mCmxm ... (2)
Multiplying (1) and (2), we get
(nC0 + nC1x + nC2x2 +...+ nCrxr +...+ nCnxn)x
(mC0 + mC1x + mC2x2 +...+ mCrxr +...+ mCmxm) = (1+x)m+n
= m+nC0 + m+2C1x + m+nC2x2 +...+ m+nCrxr +...+ m+nCm+nxm+n
Equating the coefficient of xr, we get
mCr nC0 + mCr1 nC1 + mCr2 nC2 +...+ mC1 nCr1 + mC0 nCr = m+nCr
(b) If one series has constant lower suffices and other has varying lower suffices
Illustration:
Prove that nC0.2nCn  nC12n2Cn + nC2/2n4Cn ...= 2n.
Solution:
nC0.2nCn  nC12n2Cn + nC2/2n4Cn ...
= coefficient of xn in[nC0(1+x)2n  nC1(1+x)2n2+nC2(1+x)2n4  ...]
= coefficient of xn in
[nC0((1+x)2)n  nC1((1+x)2)n1 + nC2((1+x)2)n2  ...]

= coefficient of xn in [(1+x)2  1]n
= coefficient of xn in (2x+x2)n = coefficient of xn in xn (2+x)n = 2n.
1. Divisibility problems:
Let (1 + x)n = 1 + nC1x + nC2x2 +...+ nCnxn.
In any divisibility problem, we have to identify x and n. The number by which division is to be made can be, x, x2 or x3, but the number in the base is always expressed in form of (1+x).
Illustration:
Find the remainder when 7103 is divided by 24.
Solution:
7103 = 7(50  1)51 = 7(5051  51C1 5050 + 51C2 5049  ...  1)
= 7(5051  51C15050 +...+ 51C5050)  7  18 + 18
= 7(5051  51C15050 +...+ 51C5050)  25 + 18
=> remainder is 18.
2. Questions involving the greatest integer function.
The questions generally involves working with binomial expression on surds.
Illustration:
If I is the integral part and f is the fraction part of (2 + √3 )n then prove that (l + f)(1 f) = 1. Also prove that I is an odd integer.
Solution:
(2 + √3)n, l + f where I is an integer and 0 , f < 1. We have to show that I is odd and that (l + f)(1  f) = 1
Here note that (2 + √3)n (2 + √3)n =(4  3)n = 1
.·.(2 + √3)n(2 + √3)n=1. It is thus required to prove that (2 + √3)n=1f

But, (2 + √3)n + (2 + √3)n = [2n  C1.2n1.√3 + C22n2.(√3)2  ...]
+ [2n  C1.2n1.√3 + C22n2.(√3)2  ...]
=2[2n  C2.2n2.3 + C42n4.32  ...] = even integer
.·. Now 0 < (2  √3) < 1
.·. 0 < (2  √3)n < 1
.·. If (2  √3)n = f' then l + f + f' = Even
Now ) < f < 1 and 0 < f' < 1 ... (1)
.·. f + f' = integer
(1) and (2) imply that f + f' = 1(·.· 0 < f + f' < 2)
.·. I is odd and f' = 1  f
=> (I + f)(1  f) = 1.
Binomial expression with non positive exponent
The characteristics discussed hitherto were confined to positive integer n. If n takes any other value, then binomial theorem is written as:
Say, we have to find out (x+a)n n I+
If a > 1 then it is written as an (a + (x/a))n. This is expanded as
an (1+(x/n))n = an[1+n(x/a) + (n(n1)/a) (x/a)2 + (n(n1)(n2))/3! (x/a)3 + ......∞]
Since n is not positive integer therefore the series on the right hand side will converge only for x/a < 1. Moreover, there are infinite terms in the expansion contrary to the binomial expansion for a positive integer n.
Multinomial Expression
If such a case arises, then it is not called Binomial Expansion, it is called Multinomial Expansion. If n ε N, then the general term of the multinomial expansion (x1 + x2 + x3 +...+ xk)n is (n!/a1!a2!...ak!) (x1a1,x2a2,..…xkak ), where a1 + a2 + a3+......+ ak = n and a < ai < n, I = 1, 2, 3, ...... k and the total number of terms in the expansion is n+k1Cn1.

Solved Examples
Example 1:
Find the coefficient of the independent term of x in expansion of (3x  (2/x2))15.
Solution:
The general term of (3x  (2/x2))15 is written, as Tr+1 = 15Cr (3x)15r (2/x2)r. It is independent of x if,
15  r  2r = 0 => r = 5
.·. T6 = 15C5(3)10(2)5 =  16C5 310 25.
Example 2:
If the coefficient of (2r + 4)th and (r  2)th terms in the expansion of (1+x)18are equal then find the value of r.
Solution:
The general term of (1 + x)n is Tr+1 = Crxr
Hence coefficient of (2r + 4)th term will be
T2r+4 = T2r+3+1 = 18C2r+3
and coefficient or (r  2)th term will be
Tr2 = Tr3+1 = 18Cr3.
=> 18C2r+3 = 18Cr3.
=> (2r + 3) + (r3) = 18 (·.· nCr = nCK => r = k or r + k = n)
.·. r = 6
Example 3:
If a1, a2, a3 and a4 are the coefficients of any four consecutive terms in the expansion of (1+x)n then prove that:

a1/(a1+a2) + a2/(a3+a4) = 2a2/(a2+a3)
Solution:
As a1, a2, a3 and a4 are coefficients of consecutive terms, then
Let a1 = nCr
a2 = nCr+1
a3 = nCr+2 and
a4 = nCr+3
Now a1/(a1+a2) = nCr/(nCr+nCr+1) = 1/(1+((nr)/(r+1))) = (r+1)/(n+1)
Similarly, a2/(a2+a3) = (r+3)/(n+1)
Now a3/(a3+a4) + a1/(a1+a2) = (2r+4)/(n+1)
= 2(r+1)/(n+1) = 2a2/(a2+a3) (Hence, proved)
Example 4:
Find out which one is larger 9950 + 10050 or 10150.
Solution:
Let's try to find out 10150  9950 in terms of remaining term i.e.
10150  9950 = (100+1)50  (100  1)50
= (C0.10050 + C110049 + C2.10048 +......)
= (C010050  C110049 + C210048 ......)
= 2[C1.10049 + C310047 +.........]
= 2[50.10049 + C310047 +.........]
= 10050 + 2[C310047 +............]
> 10050
=> 10150 > 9950 + 10050

Example 5:
Find the value of the greatest term in the expansion of √3(1+(1/√3))20.
Solution:
Let Tr+1 be the greatest term, then Tr < Tr+1 > Tr+2
Consider : Tr+1 > Tr
=> 20Cr (1/√3)r > 20Cr1(1/√3)r1
=> ((20)!/(20r)!r!) (1/(√3)r) > ((20)!/(21r)!(r1)!) (1/(√3)r1)
=> r < 21/(√3+1)
=> r < 7.686 ......... (i)
Similarly, considering Tr+1 > Tr+2
=> r > 6.69 .......... (ii)
From (i) and (ii), we get
r = 7
Hence greatest term = T8 = 25840/9
Example 6:
Find the coefficient of x50 in the expansion of (1+x)1000 + 2x(1+x)999 + 3x2(1+x)998 +...+ 1001x1000.
Solution:
Let S = (1 + x)1000 + 2x(1+x)999 +...+ 1000x999 (1+x) + 1001 x1000
This is an Arithmetic Geometric Series with r = x/(1+x) and d = 1.
Now (x/(1+x)) S = x(1 + x)999 + 2x2 (1 + x)998 +...+ 1000x1000 + 1000x1001/(1+x)
Subtracting we get,
(1  (x/(x+1))) S =(1+x)1000 + x(1+x)999 +...+ x1000  1001x1000/(1+x)

or S = (1+x)1001 + x(1+x)1000 + x2(1+x)999 +...+ x1000 (1+x)1001x1001
This is G.P. and sum is
S = (1+x)1002  x1002  1002x1001
So the coeff. of x50 is = 1002C50
Example 7:
Show that nCk (sin kx) cos (nk)x = 2n1 sin(nx)
Solution:
We have nCk sin kx cos (nk)x
=1/2 nCk [sin (k x + nx  kx) + sin (kx  nx + kx)]
=1/2 nCk sin n x + 1/2 nCk sin (2kx  nx)
= 1/2 sin n x nCk 1/2 [nC0 sin (nx) + nC1 sin (2n)x +...
...+ nCn1 sin (n2)x + nCn sin nx]
= 2n1 sin nx + 0 (as terms in bracket, which are equidistant, from end and beginning will cancel each other).
(Hence, proved)
Example 8:
If (15+6√6)2n+1 = P, then prove that P(1  F) = 92n+1 (where F is the fractional part of P).
Solution:
We can write
P = (15+6√6)2n+1 = I + F (Where I is integral and F is the fractional part of P)
Let F' = (15+6√6)2n+1

Note: 6√6 = 14.69 => 0 < 156√6 < 1 => 0 < (15+6√6)2n+1 < 1 => 0 < F' < 1
Now, I + F = C0 (15)2n+1 + C1(15)2n 6√6 + C2 (15)2n (6√6 )2 +...
F' = C0 (15)2n+1  C1(15)2n 6√6 + C2(15)2n1 (6√6 )2 +...
I + F + F' = 2[C0 (15)2n+1 + C2 (15)2n1 (6√6 )2 +...]
Term on R.H.S. is an even integer.
=> I + F + F' = Even integer
=> F + F' = Integer
But, 0 < F < 1 and (F is fraction part)
0 < F' < 1
=> 0 < F + F' < 2
Hence F + F' = 1
F' = (1f)
.·. P(1F) = (15 + 6√6 )2n+1 (156√6 )2n+1
= (9)2n+1 (Hence, proved)
Example 9:
Using ∫01(tx+ax)n dx,prove that ∫01xk (1x)(nk) dx=[ nCk (n+1)](1)
Solution:
The given integral can easily be evaluated, as follows:

I = ∫01(tx+ax)n dx
= ∫01((t1)x+1)n dx
= [((t1)x+1)(n+1)/((n+1)(t1))]01
=(t(n+1)1)/(n+1)(t1)=1/(1+n) [1 + t + t2 +...+ tk +...+ tn] ...... (i)
Also, I = ∫01(tx+(1x))n dx
=∫01 nCk (1x)(nK) tk xk dx ...... (ii)
Comparing the coefficient of tk from (i) and (ii), we get,
∫01 nCk. xk (1x)nk dx=1/(n+1)
=> ∫01xk (1x)nk dx = 1/( nCk (n+1))
(Hence, proved)
Example 10:
Prove that (3)r1 3nC2r1= where k = 3n/2 and n is an even positive integer.
Solution:
Since n is even integer, let n = 2m,
k = 3n/2 = 6m/2 =3m
The summation becomes,
S =
Now, (1+x)6m = 6mC0 + 6mC1 x + 6mC2 x2 +...
...+ 6mC6m1 x6m1 + 6mC6m x6m ...... (i)
(1x)6m = 6mC0 + 6mC1 (x) + 6mC2 (x)2 +...
...+ 6mC6m1 (x)6m1 + 6mC6m (x)6m ...... (ii)
=> (1+x)6m  (1x)6m = 2[6mC1x + 6mC3x2 +...+ 6mC6m1x6m2]

((1+x)6m  (1x)6m)/2x = 6mC1 + 6mC3 x2 +......+ 6mC6m1 x6m  2
Let x2 = y
=>((1 + √y)6m(1√y)6m)/2√y=6mC1+6mC3y + 6mC3 y+......6mC6m1 y3m1
With y = 3, RHS becomes = S.
LHS =
(Hence, proved)
Example 11:
If k and n are two positive integers and
Sk = 1k + 2k +.........+ nk
Then show that
m+1C1S1 + m+1C2S2 +......+ m+1CmSm = (1+n)m+1  (1+n)
Solution:
We have,
(1 + p)m+1 = m+1C0 + m+1C1 p + m+1C2 p2 + m+1Cm+1 pm+1
Putting p = 1, 2, 3, ........., n
=>2m+1  1 = m+1C0 + m+1C1(1) + m+1C2(1)2 +...+ m+1Cm(1)m
3m+1  2m+1 = m+1C0 + m+1C1(2) + m+1C2(2)2 +...+ m+1Cm(2)m
4m+1  3m+1 = m+1C0 + m+1C1(3) + m+1C2(3)2 +...+ m+1Cm(3)m

(1+n)m+1nm+1 = m+1C0(n) + m+1C2 Σn + m+1C2 Σ(n)2 +...+ m+1Cm∑(n)m.
Adding all these terms, we get
(1+n)m+1 1 = m+1C0(n) + m+1C1S1 + m+1C2S2 +...+ m+1CmSm
=>m+1C1S1 + m+1C2S2 +...+ m+1CmSm = (1+n)m+1  (1+n)
(Hence, proved)
Example 12:
Given that Sn = 1 + q + q2 +....+ qn
and Pn = 1 + ((q+1)/2) + ((q+1)/2)2 + ....+((q+1)/2)n
show that n+1C1 + n+1C2S1 + n+1C2S2 +...+ n+1Cn+1 Sn = 2n pn.
Solution:
Both Sn and Pn are geometric series
Sn = 1 + q + q2 +......qn = 1qn1/1q ...... (i)
Pn = 1 + ((q+1)/2) + ((q+1)/2)2 +.....+((q+1)/2)n
=(1((q+1)/2)n+1)/(1(q+1)/2)=1/2n(2n+1(q+1)n+1)/(1q) ...... (ii)
n+1Cr+1Sr = n+1Cr+1 ((1qr+1)/(1q)) = n+1Cr+1 ((1/(1q))  (qr+1/(1q)))
= (1/(1q))n+1Cr+1  (1/(1q)) n+1Cr+1 qr+1
=> n+1Cr+1 Sr=(1/(1q)) n+1Cr+1 (1/(1q)) n+1Cr+1qr+1
=> ∑n+1Cr+1Sr = 2npn (Hence, proved)
Example 13:
Show that if Cr is the coefficient of xr in (1+x)n, then
Cr3 = Coefficient of xn yn in the expression of ((1+x)(1+y)(x+y))n

Solution:
(1+x)n = nC0 + nC1 x + nC2 x2 +...+ nCn xn ...... (1)
(1+y)n = nC0 + nC1 y + nC2 y2 +...+ nCn yn ...... (2)
(x+y)n = nC0 xn + nC1 xn1 y + nC2 xn2 +...+ nCn yn ....... (3)
writing (1+y)n as (y+1)n and expanding it in decreasing powers of y.
(y+1)n = nC0yn + nC1yn1 + nC2yn2 +...+ nCn ....... (4)
multiplying (1), (3) and (4) we get,
(1+x)n (y+1)n (x+y)n
= (nC0 + nC1x + nC2x2 +...+ nCnxn) × (nC0yn + nC1yn1 +...+ nCn)
× (nC0xn + nC1xn1y +...+ nCnyn)
Now, coefficient of xnyn in RHS = nC03 + nC13 + nC23 +... nCn3
= coefficient of xnyn in LHS
= coefficient of xnyn in {(1+x)(1+y)(x+y)}n
(Hence, proved)
Example 14:
If (1+x)n = C0 + C1x + C2x2 +...+ Cnxn (nεN)
Then show that k3 [Ck/Ck1] = 1/12 (n)(n+1)2(n+2)
Solution:
Cr/Cr1 = (n!/(nr)!r!) (((r1)!(nr+1)!)/n!) = (((n+1)/r)1)
r3(Cr/Cr1) = ((n+1)/r1) r3 = (n+1)r2  r3
=
(n+1)n(n+1)(2n+1)/6(n2 (n+1)2)/4

= n(n+1)/12 [4n2 + 6n + 2  3n2  3n] = n(n+1)(n2+3n+2)/12
n(n+1)(n+1)(n+2)/12=(n(n+1)2 (n+2))/12
(Hence, proved)
Example 15:
If (1+x)n = C0 + C1 x + C2 x2 +...+ Cn xn then show that
(i) ∑0≤i

= 2n ∑0≤i