Question

The sum of $n$ terms of the following series $1.2+2.3+3.4+4.5+...$shall be

• A. $n^3$
• B. $\frac{n(n+1)(n+2)}{3}$
• C. $\frac{n(n+1)(n+2)}{6}$
• D. $\frac{n(n+1)(2n+1)}{3}$

Answer 1

The correct option is B

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

Explanation for the correct option:

Find the required sum.

Given: series $1.2+2.3+3.4+4.5+...n$terms

Here $t_n=n(n+1)$

$=n^2+n$

Sum,$S=\sum t_n$

$$\begin{gathered} =\sum (n^2+n) \\ =\sum n^2+\sum n \\ =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \\ =\frac{n(n+1)}{2}\left(\frac{(2n+1)}{3}+1\right) \\ =\frac{n(n+1)}{2}\times \frac{2n+4}{3} \\ =\frac{2n(n+1)(n+2)}{6} \\ =\frac{n(n+1)(n+2)}{3} \end{gathered}$$

Therefore, the sum of $n$ terms of the following series $1.2+2.3+3.4+4.5+...$shall be$\frac{n(n+1)(n+2)}{3}$.

Hence, option (B) is the correct answer.