Question

If $^nC_1 + 2\, ^nC_2 + .... + n\, ^nC_n = 2n^2$, then $n =$

• A. 4
• B. 2
• C. 1
• D. 8

Answer 1

Answer: (A)

4

Answer 2

$^{n}C_{1}+2 ^{n}C_{2}+....+n ^{n}C_{n}=2n^{2}$
$\Rightarrow \:\:\: \displaystyle\sum_{k=1}^{n} k \, ^nC_k = 2n^2$
$\Rightarrow \:\:\: \displaystyle\sum_{k=1}^{n} \frac{k(n!)}{k!(n-k)!} = 2n^2$
$\Rightarrow \:\:\: \displaystyle\sum_{k=1}^{n} \frac{k \, n(n-1)!}{k(k -1)!(n-k)!} = 2n^2$
$\Rightarrow \:\:\: n\displaystyle\sum_{k=1}^{n} \frac{(n-1)!}{(k-1)!(n-k)!} = 2n^2$
$\Rightarrow \:\:\: \displaystyle\sum_{k=1}^{n} \,^{n -1}C_{k - 1}= 2n$
$\Rightarrow \:\:\: 2^{n -1} = 2n$
$n = 4$ satisfy the above equality
Hence, $n = 4$.