If $\mathrm{n} \geq 2$ is a positive integer, then the sum of the series ${ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$ is:

(A) $\frac{\mathrm{n}(\mathrm{n}+1)^{2}(\mathrm{n}+2)}{12}$

(D) $\frac{n(2 n+1)(3 n+1)}{6}$

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Binomial Theorem

n \geq 2

is a positive integer, then the sum of the series

^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{n} C_{2})

is:

Answer: (B)

By the identity

r = 2 \sum n (2 r) = (3 n + 1)

, the series equals

(2 n + 1) + 2 (3 n + 1)

. Expanding both terms,

(2 n + 1) + 2 (3 n + 1) = \frac{n ( n + 1 )}{2} + \frac{n ( n + 1 ) ( n - 1 )}{3} = \frac{n ( n + 1 )}{6} [3 + 2 (n - 1)] .

The bracket is

2 n + 1

, so the sum of the series is

\frac{n ( n + 1 ) ( 2 n + 1 )}{6}

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option