RBSE Solutions for Class 11 Maths Chapter 7 Binomial Theorem Ex 7.3

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.3

Question 1.
If C0, C1, C2,… Cn are coefficients of expansion (1 + x)n then find the value of :
8C1 + 8C2 + 8C3 + … + 8C8
8C1+ 8C3 + 8C5 + 8C7
Solution:
(i) 8C1 + 8C2 + 8C3 + … + 8C8 We know that
nC0 + nC1+ nC2 + … + nCn = 2n
Putting n = 8
8C0 + 8C1 + 8C2 + …. + 8C8 = 28
⇒ 1 +8C1 + 8C2 + 8C3 + … + 8C8 = 28
8C1 + 8C2 + …. + 8C8 = 28 – 1 = 255.

(ii) 8C1 + 8C3 + 8C5 + 8C7
We know that
8C1 + 8C3 + 8C5 +… = 2n-1
Putting n = 8
8C1 + 8C3 + 8C5 + 8C7 = 28 -1 = 27 = 128.

Question 2.
If C0, C1, C2,… Cn are coefficients of expansion (1 + x)then prove that:
C0 + 3.C1 + 5.C2 + … + (2n + 1). Cn = (n + 1)2n.
Solution:
L.H.S.
= nC0 + 3.C1 + 5.C2 + … + (2n + 1) nCn
= nC0 + (2 + 1) nC1 + (4 + 1) nC2 + … (2n + 1) nCn
= nC0 + (2. nC1 + nC1 + (4. nC2 + nC2) + …  (2n.nCn + nCn)
= 2. nC1 + 4 nC2 + 6 nC3 + … + 2n nCn
+ (nC0 + nC1 + nC2 + nC3 + …nCn)
= 2[nC1 + 2 nC2 + 3.nC3 +… + nCn] + (1 + 1)n

Question 3.
CC2 + CC3 + CC4 + … + Cn _ 2Cn= $$\frac { (2n)! }{ (n-2)!(n+2)! }$$
Solution:
From Binomial theorem
(1 + x)n = C0 + C1x + C2x2 + … + Crxr  + … + Cnxn … (i)
Again
(x + 1)n = C0xn + C1xn-1 + C2xn-2 + …+ Crxnr+ …+ Cn-1 x + Cn …(ii)
Multiplying equation (i) and (ii) we get
(1 + x)2n = [C0 + C1x + C2x2 + … + Crxr  + … + Cnxn]
× [C0xn + C1xn-1 + C2xn-2 + …+ Crxnr+ …+ Cn-1 x + Cn]
Comparing cofficient of xnr in both sides, we get
C0Cr + C1Cr+1+ C2Cr +2 + … + Cn _ rCn

Question 4.
C0 + 2C1 + 4C2 + 6C3 + … + 2nCn = 1 + n2n.
Solution:
L.H.S.
= C0 + 2C1 + 4C2 + 6C3 + … + 2nCn
= C0 + 2[C1 + 2C2 + 3C3 + … + nCn]

Question 5.

Solution:
L.H.S.

Question 6.
If expansion of (1 + x – 2x2)6 is denoted by 1 + a1x+ a2x2 + a3x3 + … + a12x12 then prove that a2 + a4 + a6 + … + a12 = 31.
Solution :
Given expansion is :
(1 + x – 2x2)6 = 1 + a1x+ a2x2 + a3x3 + … + a12x12 … (i)
Putting x = – 1 in equation (i)
{1 + 1 – 2(1)2}6 = 1+ a1 + a2 + a3 + … + a12
⇒ (2 – 2)6 = 1 + a1 + a2 + a3 + … + a12
⇒ 1 + a1 + a2 + a3 + … + a12 = 0 … (ii)
Putting x = – 1 in equation (i)
{(1 – 1 – 2(-1)2}6 = 1 – a1 + a2 + a3 + … + a12
⇒ (-2)6 = 1 – a1 + a2 + a3 + … + a12
⇒ 1- a1 + a2 + a3 + … + a12 = 64       … (iii)
Adding equation (ii) and (iii) we get
⇒ 2 + 2a2 + 2a4 + … + a12 = 64
⇒   2(1 + a2 + a4 + … + a12 = 64
⇒  1 + a2 + a4 + … + a12 = 64/2 =32
⇒ a2 + a4 + a6 … + a12 = 32 – 1 = 31
Hence, a2 + a4 + a6 … + a12 =31.
Hence proved.