Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ... - Mathematics

1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

#### Solution

Let $T_n$ be the nth term of the given series.
Thus, we have: $T_n = n\left( n + 1 \right) = n^2 + n$

Now, let $S_n$ be the sum of n terms of the given series.

Thus, we have: $S_n = \sum^n_{k = 1} T_k$

$\Rightarrow S_n = \sum^n_{k = 1} \left( k^2 + k \right)$

$\Rightarrow S_n = \sum^n_{k = 1} k^2 + \sum^n_{k = 1} k$

$\Rightarrow S_n = \left( \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{n\left( n + 1 \right)}{2} \right)$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{2n + 1}{3} + 1 \right)$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{2n + 4}{3} \right)$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 4 \right)}{6}$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)\left( n + 2 \right)}{3}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.1 | Q 6 | Page 10