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Continue Shopping ```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.

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 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)!)

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 = 2nCn-r = (2n)/((n-r)!(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 + C1xn-2 +...+ Cn,                                            ... (2)

(C0 + C1x + C2x2 +...+ Cnxn)(C0xn + C1xn-1 + C2xn-2 +...+ 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 + 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.

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