If n ⩾ 2 is a positive integer, then the sum of the series C2 n + 1 + 2(C2 2 + C2 3 + C2 4 + + C2 n) is

Correct answer: B. (n(n + 1)(2n + 1))/(6) C2 n + 1 + 2(C2 2 + C2 3 + C2 4 + + C2 n) C2 n + 1 + 2(C3 3 + C2 3 + C2 4 + + C2 n) use Cr + 1 n + Cr n = Cr n + 1 = C2 n + 1 + 2(C3 4 + C2 4 + C3 5 + + C2 n) = C2 n + 1 + 2(C3 5 + C2 5 + + C2 n) = C2 n + 1 + 2(C3 n + C2 n) = C2 n + 1 + 2 · C3 n + 1 = ((n + 1)n)/(2) + 2 · ((n + 1)(n)(n - 1))/(2 . 3) = (n(n + 1)(2n + 1))/(6)

JEE Main 202124 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Binomial Theorem

If n ⩾ 2 is a positive integer, then the sum of the series C2 n + 1...

n ⩾ 2

is a positive integer, then the sum of the series

C2 n + 1 + 2\left(C2 2 + C2 3 + C2 4 + \ldots + C2 n\right)

Options

A $\dfrac{n\left(n - 1\right)\left(2n + 1\right)}{6}$
B $\dfrac{n\left(n + 1\right)\left(2n + 1\right)}{6}$
C $\dfrac{n{\left(n + 1\right)}^{2}\left(n + 2\right)}{12}$
D $\dfrac{n\left(2n + 1\right)\left(3n + 1\right)}{6}$

Correct answer

B. $\dfrac{n\left(n + 1\right)\left(2n + 1\right)}{6}$

Step-by-step solution

$C2 n + 1 + 2\left(C2 2 + C2 3 + C2 4 + \ldots + C2 n\right)$

$C2 n + 1 + 2\left(C3 3 + C2 3 + C2 4 + \ldots + C2 n\right)$

{use $Cr + 1 n + Cr n = Cr n + 1$ }

$= C2 n + 1 + 2\left(C3 4 + C2 4 + C3 5 + \ldots + C2 n\right)$

$= C2 n + 1 + 2\left(C3 5 + C2 5 + \ldots + C2 n\right)$

$= C2 n + 1 + 2\left(C3 n + C2 n\right)$

$= C2 n + 1 + 2 \cdot C3 n + 1$

$= \dfrac{\left(n + 1\right)n}{2} + 2 \cdot \dfrac{\left(n + 1\right)\left(n\right)\left(n - 1\right)}{2 . 3}$

$= \dfrac{n\left(n + 1\right)\left(2n + 1\right)}{6}$

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