Binomial theorem for any index

BINOMIAL THEOREM FOR ANY INDEXLet n be a rational number and x be a real number such that |x| < 1 , then

(1+x)n = 1+ nx + n(n−1)2!

x2 + . . . +n (n−1 ) (n−2 )……(n−r+1)

r ! xr + ... terms upto

∞ ......(1)

Observations:

o General term of the series (1+x)-n = Tr+1 = (-1)r n (n+1 ) (n+2 )……(n+r−1)r !

xr

o General term of the series (1-x)-n = Tr+1 = n (n+1 ) (n+2 )……(n+r−1)

r ! xr

o If first term is not 1, then make first term unity in the following way:

(a+ x)n = an(1+xa)

n if < |xa |< 1

IMPORTANT EXPANSIONS· (1+ x)-1 = 1- x +x2 –x3 + . . . + (-1)rxr+. . .· (1 - x)-1 = 1+ x +x2 +x3 + . . .+ xr + . . .· (1+ x)-2 = 1- 2x +3x2 –4x3+ . . .+ (-1)r(r+1)xr+. . .· (1 - x)-2 = 1+ 2x +3x2 +4x3 + . . .+ (r+1)xr+. . .

· (1+x)-3 = 1- 3x +6x2 –10x3 +. . .+ (-1)r (r+1 )(r+2)r !

xr+. . .

· (1-x)-3 = 1+ 3x +6x2 +10x3 + . . .+ (r+1 )(r+2)

r ! xr+. . .

· In general coefficient of xr in (1 – x)– n is n + r –1Cr . · (1 – x)–p/q = 1 + p/1! (x/q) + p(p+q)/2! (x/q)2 + ……. From (1)· (1 + x)–p/q = 1 – p/1! (x/q) + p(p+q)/2! (x/q)2 – …….· (1 + x)p/q = 1 + p/1! (x/q) + p(p–q)/2! (x/q)2 + …….· (1 – x)p/q = 1 – p/1! (x/q) + p(p–q)/2! (x/q)2 – …….

Solved Example 1: If –1 < x < 1, show that (1 –x)-2 = 1 + 2x + 3x2 + 4x3 + …..to ∞.Solution: We know that if n is a negative integer or fraction

(1+x)n= 1 + Provided –1 < x < 1

Putting n = –2 and –x in place of x, we get

(1+x)2 = 1 + = 1 + 2x + 3x2 + 4x3 + … to ∞.

Properties Of Binomial Expansion1. 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 = nCn-r = n!/r!(n-r)! .(coeffts. Of terms equidistant from beginning and the end are equal)

2. The general term in the expansion of (a + x)n is (r + 1)th term given as

Tr+1 =nCran-r xr.

Similarly the general term in the expansion of (x + a)n is given as Tr+1 = nCrxn-r 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 = nCn-r.

Note: Here we are using n

Cr + n

Cr-1 = n+1

Cr this concept will be discussed later in this chapter.

Also, we have replace m

C0 by m+1

C0 because numerical value of both is same i.e. 1. Similarly we

replace m

Cm by m+1

Cm+1.

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 +...... = 2n-1.

nCr =

nr

. n-1Cr-1 = nr

n−1r−1

. n-2Cr-2 and so on.....

crn

cr−1n =n−r+1

r , r = 1,2,3.....

EXAMPLE: If C0 , C1 , C2 ... Cn denote the binomial coeffts. In the expansion of (1+x)n , prove that:

(i)c1c0

+ 2c2c1

+ 3c3c2

+.........+ncncn−1

= n(n+1)2

(ii) (C0+ C1)( C1+ C2)( C2+ C3)..........( Cn-1+ Cn) = co c1 c2…………cn−1 ¿¿

SOLUTION: use

crn

cr−1n =n−r+1

r , r = 1,2,3.....

r. crn

cr−1n = n-r+1, r=1,2,3....n

∑r=1

n

rcrcr−1

= ∑r=1

n

(n−r+1) = n+(n-1)+.....+3+2+1

(ii) (c0+c1c0

) (c1+c2¿¿c1

).........(cn−1+cn ¿¿cn−1

) = (n+1)n

n !

¿ )(1+c2c1

).........(1+cncn−1

) (either can take only L.H.S. &mult. And

divide by c0.c1.c2.....cn ) now use above result.

Illustration:

If (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, then prove that C0 + (Co + C1) + (C0 + C1+ C2) + ... (C0 + C1 + C2 +...+ Cn-1) = n2n-1 (where n is an even integer.

Solution:

C0 + (C0 + C1) + (C0 + C1 + C2) +... (C0 + C1 + C2 +...+ Cn-1)

= C0+(C0 + C1 + C2 +...+ Cn-1)+(C0 + C1)+(C0 + C1 + C2 +...+ Cn-2)+...

= (C0 + C1 + C2 +...+ Cn)+ (C0 + C1 + C2 +...+ Cn)+... n/2 times

= (n/2)2n = n . 2n-1

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; (1-x)n = nC0 - nC1 x + nC2 x2 -...+ (-1)n nCn xn.

Now, adding both expansions, we get,

(1 + x)n + (1-x)n = 2[nC0 + nC2 x2 + nC4 x4 +.........]

Put x = 1

=> (2n+0)/2 = C0 + C2 + C4 +......

or C0 + C2 + C4 +......= 2n-1

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)n-1 = 1 C1 x1-1 + 2 C2 x2-1 +...+ n Cn xn-1

Put x = 1, we get, n 2n-1 = + 1 C1 + 2 C2 +...+ n Cn.

Or, ∑nr=0 r Cr = n 2n-1, which is the answer.

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 + C4-C6 +.........

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 C0-C1 + C2-C3 +... 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 + (1-i)n)/2 ] [C0 - C2 + C4 + ......]

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.

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!) (x1

a1,x2a2,..…xk

ak ), 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+k-1Cn-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 nCn-1 + nCn-2 +...+ nCn nC0

i.e. 0 + n = 1 + (n-1) = 2 + (n-2) = 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 +...+Cn-r Cn =

2n!n-r!n+r!(Similarily we can prove that C0C1 + C1C2 + C2C3 +...+ Cn-1 Cn =

(2n! )(n−1 ) ! (n+1 ) !

Solution:

We have,

C0 + C1x + C2x2 + ... + Cnxn = (1+x)n ... (1)

Also C0xn + C2xn-2 +...+ Cn = (x+1)n ... (2)

Multiplying (1) and (2), we get

(C0 + C1x2 +...+ Cnxn)(C0xn + C1xn-2 + C2xn-2 +...+ Cn) = (1+x)2n ... (3)

Equating coefficient of xn-r from both sides of (3), we get

C0Cr + C1Cr+1 + C2Cr+2 +...+ Cn-rCn =

Cn−r2n =

(2n! )(n−r )! (n+r )!

Illustration:

(i) Prove that Co2 + C1

2 +...+ Cn2 = (2n)!/n!n! .

(ii) prove that Co2 - C1

2 +c22...+(-1)n Cn

2 ={0 , if n is odd(−1)n2

if n is even

Solution:

(i) Since

(1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, ... (1)

(x + 1)n = C0xn + C1xn-2 +...+ Cn, ... (2)

(C0 + C1x + C2x2 +...+ Cnxn)(C0xn + C1xn-1 + C2xn-2 +...+ Cn)=(1+x)2n.

Equating coefficient of xn, we get

C02 + C1

2 + C22 +...+ Cn

2 = 2nCn = (2n)!/n!n! .

(ii) (x+1)n(1-x)n = (C0xn + C1x(n-1) + C2xn-2 +...+ Cn-1x+Cn )( C0 - C1x + C2x2 -C3x3+C4x4...+(-1)n Cnxn )=

(1-x2)n = C0 - C1x2 + C2x4 -C3x6+...+(-1)n Cnx2n

Equating the coeffts of xn on both sides

Co2 - C1

2 +c22...+(-1)n Cn

2 = 0, if n is odd

If n is even , suppose (r+1)th term in the expansion of (1-x2)n contains xn

, we have Tr+1 = crn (-1)r x2r it implies 2r =n i.e., r=n/2 hence proved.

Example: Find the coeffts. Of x4 in the expansion of (1+x)n(1-x)n.

Deduce that C2 = C0C4 - C1C3 + C2C2 – C3 C1+ C4C0

SOLUTION: (1+x)n(1-x)n = (C0 + C1x + C2x2 +C3x3+C4x4...+ Cnxn )( C0 - C1x + C2x2 -C3x3+C4x4...+(-1)n Cnxn )

Coeffts of x4 on R.H.S. = C0C4-C1C3+C2C2-C3C1+C4C0

(1-x2)n = C0 - C1x2 + C2x4 -C3x6+...+(-1)n Cnx2n

Illustration:

If (1+x)n = , then prove that

mCr nC0 + mCr-1 nC1 + mCr-2 nC2 +...+ mC1 nCr-1 + 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 + mCr-1 nC1 + mCr-2 nC2 +...+ mC1 nCr-1 + mC0 nCr = m+nCr

(b) If one series has constant lower suffices and other has varying lower suffices

Illustration:

Prove that nC0.2nCn - nC12n-2Cn + nC2.2n-4Cn -...= 2n.

Solution:

nC0.2nCn - nC12n-2Cn + nC2/2n-4Cn -...

= coefficient of xn in[nC0(1+x)2n - nC1(1+x)2n-2+nC2(1+x)2n-4 - ...]

= coefficient of xn in

[nC0((1+x)2)n - nC1((1+x)2)n-1 + nC2((1+x)2)n-2 - ...]

= coefficient of xn in [(1+x)2 - 1]n

= coefficient of xn in (2x+x2)n = co-efficient of xn in xn (2+x)n = 2n.

Some Important Results

1. Differentiating (1+x)n = C0 + C1x + C2x2 +...+ Cnxn of both sides we have,

n(1 + x)n-1 = C1 + 2C2x + 3C3x2 +...+ nCnxn-1. ... (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)n-1 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+1-1)/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 = n2n-1.

Solution:

Method (i) : rth term of the given series

rth term of the given series, tr = nCr

tr = r × n/r × n-1Cr-1 = n × n-1Cr-1 (because nCr =n/r .n-1Cr-1)

Sum of the series =

Put x = 1 in the expansion of (1 + x)n-1, so that

(n-1C0 + n-1C1 +...+ n-1Cn-1) = 2n-1

=> = n.2n-1.

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)n-1 = C1 + 2C2x + 3C3 x2 +...+ n Cnxn-1. ... (2)

Putting x = 1 in (2), we have, n 2n-1 = C1 + 2C2 +....+ nCn. ... (3)

Illustration:

If (1+ x)n = xr then prove that C0 + 2.C1 + 3.C2+...+(n+1)Cn

=2n-1(n+2).

Solution:

Method (i): rth term of the given series

rth term of the given series

tr = nCr-1 = [(r-1) + 1]. nCr-1

= (r-1) nCr-1 + nCr-1 = n. n-1Cr-2 + nCr-1 (because nCr-1 =n/(r-1) . n-1Cr-2)

Sum of the series =

= n[n-1C0 + n-1C1 +...+ n-1Cn-1]+[nC0 + nC1 +...+ nCn] = n.2n-1 + 2n = 2n-1 (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)n-1 x = C0x + 2C1x2 +...+ (n+1)Cnxn ... (3)

Putting x = 1 in (3), we get

2n + n.2n-1 = C0 + 2C1 + 3C2 +...+ (n+1)Cn

=> C0 + 2C1 + 3C2 +...+ (n+1)Cn = 2n-1 (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.

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.

∑r=0

n crr+1

= ∑r=0

n1n+1

.n+1r+1

cr = 1n+1 ∑r=0

nn+1r+1

cr =1n+1 [ c1

n+1 + c2n+1 +¿..... cn+1

n+1 ¿

{ use cr+1n+1 =

n+1r+1 . cr

n , and add & subtract c0n+1 }=

1r+1 [2

(n+1) -1]

Similarily we can prove C0/1 - C1/2 - C2/3 +......+(-1)n Cn/n+1 = 1/(n+1)

L.H.S. = ∑r=0

n

(−1)rcrr+1

=1n+1

∑r=0

n

(−1)r n+1r+1

cr=1n+1

[ c1n+1 − c2

n+1 −¿....

(−1)n cn+1n+1 ¿

Or another method

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,

.

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/Cr-1 = (n-r+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 = (n-1)/2 or (n+1)/2 and the coefficient itself will be nC(n+1)/2 or nC(n-1)/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...(2n-1).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...(2n-1).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 (3-2x)9 when x = 1.

Solution:

Tr+1/Tr = ((9-r+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 = ((n-r+1)/r)(3x/2) = ((10-r)/r)(3x/2), (where x = 3/2)

= ((10-r)/r)(3x/2)(3/2) = ((10-r)/r).9/4 = (90-9r)/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/(21-r)).√3.

Now Tr/Tr+1 > 1 => r > 21/(√3+1) ...... (1)

Now again Tr/Tr+1 = (r/(21-r)) √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).

Illustration:

Find the first negative term in the expansion of (1+x)7/2.

Solution: Tr+1 =

72.( 72−1)(72−2)… ..( 72−r+1) xr

r !

this will be first

negative term if

( 72−r+1) <0 it implies r>9/2 or r≥5 , hence 6th term is the first

negative term = -7/256 x5 ( by putting r=5).

Illustration:

If (1-x)-n = a0+a1x+a2x2+a3x3+......., find the value of a0+a1+a2+......+an.

Solution: use (1-x)-n(1-x)-1=(a0+a1x+a2x2+.....+anxn+...)(1+x+x2....+xn+....)

Coeffts. Of xn in (1-x)-(n+1) = a0+a1+a2+......+an

Tr+1=−(n+1 ) (−(n+1 )−1 ) (−(n+1 )−2 )…………. {− (n+1 )−(r−1 ) } .(−x)r

r ! =

(−1)2 r (n+1 ) (n+2 )…….(n+r ) xr

r !

Put r=n then coeffts. Of xn in (1-x)-(n+1)= a0+a1+a2+......+an =(n+1 ) (n+2 )…….(n+n)

n! = (2n)!/(n!)2

Illustration:

Find the coefft. Of x6 in (1+x+x2)-3

Solution: (1+x+x2)-3 ={(1−x )(1+x+ x2)

(1−x )}-3 = ¿=(1-x)3(1-x3)-3

Coefft. Of x6 is 6-3=3.

Particular Cases

We have (a + x)n = an + nC1 an-1 x + nC2 an-2 xr +...+ xn. ...... (1)

(i) Putting x = -x in (1), we get

(a-x)n = an - nC1 an-1x + nC2 an-2 x2 - nC2 an-3 x3 +...+ (-1)r nCr an-r 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

(1-x)n=nC0-nC1 x + nC2 x2-nC3x3 +... (-1)r nCr xr +... (-1)n nCnxn ... (B)

Tips to Remember

(a) Tr/Tr+1 = ((n-r+1)/r ).(x/a) for the binomial expansion of (a + x)n.

(b) (n+1)Cr = nCr + nCr-1.

(c) r nCr = n n-1Cr-1

(d)

(e) When n is even,

(x + a)n + (x - a)n = 2(xn + nC2 xn-2 a2 + nC4 xn-4 a4 +...+ nCn an).

When n is odd,

(x + a)n + (x - a)n = 2(xn + nC2 xn-2 a2 +...+ nCn-1 x an-1).

When n is even

(x + a)n - (x - a)n = 2(nC1 xn-1 a + nC3 xn-3 a3 +...+ nCn-1 x an-1).

When n is odd

(x + a)n - (x - a)n = 2(nC1 xn-1 a + nC3 xn-3 a3 +...+ nCn an).

Binomial Theorem(Approximation)

Illustration:

If p is nearly to q and n>1, show that (n+1 ) p+ (n−1 )q(n−1 ) p+(n+1 )q

=( pq)1n ,hence

find ( 99101

)16 .

Solution: let p = q+h, where h is very small that its square & higher powers may be neglected. Then ,L.H.S.

(n+1 )(q+h)+(n−1 )q(n−1 )(q+h)+(n+1 )q =

2nq+(n+1 )h2nq+(n−1 )h =

1+((n+1)2nq

)h

1+((n−1)2nq

)h =

{1+( (n+1 )2nq )h } {1+( (n−1 )

2nq )h} -1

={1+( (n+1 )2nq )h}{ 1−(

(n−1)2nq

)h}={1+h/(nq)} [expanding neglecting

higher power of h]R.H.S.

( pq)1n = ( q+h

q)1n =(1+ h

q)1n = 1+h/(nq) [ same reason]

Put p=99, q= 101 and n=6 in above result , we get ( 99101

)16 =

599/601.

Illustration:

Assuming x to be so small that x2 and higher power of x can be neglected, show that:

(1+ 34x)

−4

(16−3 x)12

(8+x)23

is approx. Equal to 1- (305x/96).

Solution: (1+ 3

4x)

−4

(16−3 x)12

(8+x)23

= (1+ 3

4x)

−4

(16)12 (1−3 x

16)12

823 (1+ x

8)23

= (1+ 34x)

−4

(1−3 x16

)12 (1+ x

8)−23 =(1-3x)(1-3x/32)(1-x/12) [by using

(1+x)n =1+nx] =(1-3x-3x/32)(1-x/12)= (1-99x/32-x/12)=1-305x/96. [by neglecting higher power of x]

Similarily, we can find a and b , if (1+ 3

4x)

−4

(16−3 x)12

(8+x)23

= a+bx for all

small values of x , a= 1, b= -305/96 by equating first two terms on the both sides.

SOME EXAMPLES

Example1 :

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+((n-r)/(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)

Example2 :

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

Example3 :

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 > 20Cr-1(1/√3)r-1

=> ((20)!/(20-r)!r!) (1/(√3)r) > ((20)!/(21-r)!(r-1)!) (1/(√3)r-1)

=> 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 4:

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 5:

Show that nCk (sin kx) cos (n-k)x = 2n-1 sin(nx)

Solution:

We have nCk sin kx cos (n-k)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 (2-n)x +...

...+ nCn-1 sin (n-2)x + nCn sin nx]

= 2n-1 sin nx + 0 (as terms in bracket, which are equidistant, from end and beginning will cancel each other.

Example 6:

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)2n-1 (6√6 )2 +...

I + F + F' = 2[C0 (15)2n+1 + C2 (15)2n-1 (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' = (1-f)

.·. P(1-F) = (15 + 6√6 )2n+1 (15-6√6 )2n+1

= (9)2n+1

Example 7:

Using ∫01(tx+a-x)n dx,prove that ∫0

1xk (1-x)(n-k) dx=[ nCk (n+1)](-1)

Solution:

The given integral can easily be evaluated, as follows:

I = ∫01(tx+a-x)n dx

= ∫01((t-1)x+1)n dx

= [((t-1)x+1)(n+1)/((n+1)(t-1))]01

=(t(n+1)-1)/(n+1)(t-1)=1/(1+n) [1 + t + t2 +...+ tk +...+ tn] ...... (i)

Also, I = ∫01(tx+(1-x))n dx

=∫01 nCk (1-x)(n-K) tk xk dx ......

(ii)

Comparing the coefficient of tk from (i) and (ii), we get,

∫01 nCk. xk (1-x)n-k dx=1/(n+1)

=> ∫01xk (1-x)n-k dx = 1/( nCk (n+1))

(Hence, proved)

Example 8:

Prove that (-3)r-1 3nC2r-1= 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 +...

...+ 6mC6m-1 x6m-1 + 6mC6m x6m ...... (i)

(1-x)6m = 6mC0 + 6mC1 (-x) + 6mC2 (-x)2 +...

...+ 6mC6m-1 (-x)6m-1 + 6mC6m (-x)6m ...... (ii)

=> (1+x)6m - (1-x)6m = 2[6mC1x + 6mC3x2 +...+ 6mC6m-1x6m-2]

((1+x)6m - (1-x)6m)/2x = 6mC1 + 6mC3 x2 +......+ 6mC6m-1 x6m - 2

Let x2 = y

=>((1 + √y)6m-(1-√y)6m)/2√y=6mC1+6mC3y + 6mC3 y+......6mC6m-1 y3m-1

With y = -3, RHS becomes = S.

LHS =

Example9:

If (1+x)n = C0 + C1 x + C2 x2 +...+ Cn xn then show that

(i) ∑0≤i<j≤n Ci Cj = 22n-1 -

(ii) ∑0≤i<j≤n (Ci + Cj)2 = (n - 1) 2nCn + 22n

(iii) ∑0≤i<j≤n ∑Ci Cj = n (22n-1 - 1/2 2nCn)

Solution:

(i) We have

(C0 + C1 + C2 +...+ Cn)2 = C02 + C1

2 +...+ Cn2 + 2 ∑0≤i<j≤n ∑Ci Cj

we get, (2n)2 = 2nCn + 2 ∑0≤i<j≤n ∑Ci Cj

therefore ∑0≤i<j≤n ∑Ci Cj = 22n-1 - (2n)!/2(n!)2

(Hence, proved)

(ii) ∑0≤i<j≤n ∑Ci Cj n(C02 + C1

2 + C22 +...+ Cn2) + 2∑0≤i<j≤n ∑Ci Cj

= n 2nCn + 2{22n-1 - (2n)!/2(n!)2} [from part (i)]

= n 2nCn + 22n - 2nCn

= (n-1) 2nCn + 22n

(Hence, proved)

(iii) Let ∑0≤i<j≤n ∑Ci Cj

replace i by (n - i) and j by (n - j), we have

P = ∑0≤i<j≤n ∑(2n-i0j) Cn-i Cn-j

= ∑0≤i<j≤n ∑(2n-(i+j)) Ci Cj [·.· nCn = nCn-r]

= 2n ∑0≤i<j≤n ∑Ci Cj - P

.·. ∑0≤i<j≤n ∑(i+j) Ci Cj = n[22n-1 - (2n)!/2(n!)2]

Example 10. Find the value of (0.98)-3 upto 2 decimal places.

Solution: (1-0.02)-3= [1+(-3)(-0.02)+(-3)(-3-1)/2!(-0.02)2+ ......] =[1+0.0600+0.0024] 1.0624 = 1.06 { by neglecting higher power

ofmore zeros just after decimal}.

By: --indu thakur

Binomial theorem for any index

Education

Transcript of Binomial theorem for any index