First slide

Binomial theorem for positive integral Index

Question

The sum of ${}^{n + 1}C_{3} + 2 [{}^{2}C_{2} + {}^{3}C_{2} + {}^{4}C_{2} + .... + {}^{n}C_{2}]$ is

Moderate

A

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

$\frac{n (n^{2} - 1)}{2}$
C

$\frac{n (n^{2} - 1)}{6}$
D

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

Solution

${}^{n + 1}C_{3} + 2 [{}^{2}C_{2} + {}^{3}C_{2} + {}^{4}C_{2} + .... + {}^{n}C_{2}] = {}^{n + 1}C_{3} + 2 [{}^{3}C_{3} + {}^{3}C_{2} + {}^{4}C_{2} + .... + {}^{n}C_{2}] = {}^{n + 1}C_{3} + 2 [{}^{4}C_{3} + {}^{4}C_{2} + .... + {}^{n}C_{2}] (∵ {}^{n}C_{r} + {}^{n}C_{r + 1} = {}^{n + 1}C_{r + 1}) = {}^{n + 1}C_{3} + 2 [{}^{5}C_{3} + .... + {}^{n}C_{2}] = {}^{n + 1}C_{3} + 2 [{}^{n + 1}C_{3}] = 3 {}^{n + 1}C_{3} = 3 \frac{(n + 1) (n) (n - 1)}{3 \times 2} = \frac{n (n^{2} - 1)}{2}$

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