(A) The $n^{th}$ term of the series is $a_n = n(n+1)(n+2)$.We can write $a_n = \frac{1}{4} [n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)]$.Sum $S_n = \sum_{k=1

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(A) The $n^{th}$ term of the series is $a_n = n(n+1)(n+2)$.We can write $a_n = \frac{1}{4} [n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)]$.Sum $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n \frac{1}{4} [k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)]$.This is a telescoping sum.$S_n = \frac{1}{4} [n(n+1)(n+2)(n+3) - 0(1)(2)(3)]$.$S_n = \frac{n(n+1)(n+2)(n+3)}{4}$.

Class 11 Mathematics Sequences and Series VN or Telescoping Method

Difficult

Find the sum to $n$ terms of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$

A
$\frac{n(n+1)(n+2)(n+3)}{4}$
B
$\frac{n(n+1)(n+2)(n+4)}{4}$
C
$\frac{n(n+1)(n+2)(n+5)}{4}$
D
$\frac{n(n+1)(n+3)(n+4)}{4}$

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Sequences and Series

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VN or Telescoping Method

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DifficultJEE MAIN 2024

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Easy

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The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is:

DifficultJEE MAIN 2021

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Find the sum to $n$ terms of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$

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