Find the value of nC0 4 + 42 × nC12 + ......... 4n+1n+1 × nCn
Find the value of nC0 4 + 42 × nC12 + ......... 4n+1n+1 × nCn
The correct option is D
Method - 1
Consider the expansion of (1+x)n
(1+x)n = nC0 + nC1 x + nC2 x2 + ............nCnx^n -----------(1)
We want to find nC0 4 + 42 × nC12 + .............. 4n+1n+1 × nCn
Each term is of the form
(4)r+1× nCr(r+1). If we integrate the terms in (1), we will get the terms as xr+1 ×nCrr+1 .
If we put x = 4, we will get the required sum.
Since we are integrating, we have to decide the upper and lower limits. The upper limit is 4 because we
want 4r+1 in the terms lower limit will be zero.
⇒ sum = 4∫0(1+x)n = 4∫0 nC0 + 4∫0 nC1x + ..............4∫0 nCnxn
= [(1+x)n+1]40n+1 = [nC0x]40 + [nC1x22]40.................. [nCnxn+1]40n+1
⇒ 5n+1−1n+1 = nC0 4 + 42 × nC12 + ......... 4n+1n+1 × nCn
Method 2
Each term of the sum is of the form
4r+1 × nCrr+1, where r varies from 0 to n
Consider the relation n+1Cr+1 = n+1r+1 nCr
⇒ nCrr+1 = n+1Cr+1 × 1n+1
⇒ sum = n∑r=04r+1 nCrr+1 = ∑nr=04r+1×n+1Cr+1n+1
= 1n+1 n∑r=04r+1 n+1Cr+1
= 1n+1 (4×n+1C1+n+1C2×42.....n+1Cn+14n+1)
The terms in the bracket are part of the expansion of (1+x)n+1 when x = 4. Only the first term is
missing. We will add and substract it.
⇒ sum = 1n+1 (n+1C0+n+1C14+...........−n+1C0)
= 1n+1((1+4)n+1−1)
= 5n+1−1n+1
Method - 1
Consider the expansion of (1+x)n
(1+x)n = nC0 + nC1 x + nC2 x2 + ............nCnx^n -----------(1)
We want to find nC0 4 + 42 × nC12 + .............. 4n+1n+1 × nCn
Each term is of the form
(4)r+1× nCr(r+1). If we integrate the terms in (1), we will get the terms as xr+1 ×nCrr+1 .
If we put x = 4, we will get the required sum.
Since we are integrating, we have to decide the upper and lower limits. The upper limit is 4 because we
want 4r+1 in the terms lower limit will be zero.
⇒ sum = 4∫0(1+x)n = 4∫0 nC0 + 4∫0 nC1x + ..............4∫0 nCnxn
= [(1+x)n+1]40n+1 = [nC0x]40 + [nC1x22]40.................. [nCnxn+1]40n+1
⇒ 5n+1−1n+1 = nC0 4 + 42 × nC12 + ......... 4n+1n+1 × nCn
Method 2
Each term of the sum is of the form
4r+1 × nCrr+1, where r varies from 0 to n
Consider the relation n+1Cr+1 = n+1r+1 nCr
⇒ nCrr+1 = n+1Cr+1 × 1n+1
⇒ sum = n∑r=04r+1 nCrr+1 = ∑nr=04r+1×n+1Cr+1n+1
= 1n+1 n∑r=04r+1 n+1Cr+1
= 1n+1 (4×n+1C1+n+1C2×42.....n+1Cn+14n+1)
The terms in the bracket are part of the expansion of (1+x)n+1 when x = 4. Only the first term is
missing. We will add and substract it.
⇒ sum = 1n+1 (n+1C0+n+1C14+...........−n+1C0)
= 1n+1((1+4)n+1−1)
= 5n+1−1n+1
