Question

The sum of $(n-1)$ terms of $1+(1+3)+(1+3+5)+...$is

• A. $\frac{n(n+1)(2n+1)}{6}$
• B. $n^2$
• C. $\frac{n(n-1)(2n-1)}{6}$
• D. none of these

Answer 1

The correct option is C

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

Explanation for the correct option:

Find the required sum.

Given series: $1+(1+3)+(1+3+5)+...$

$t_n=(1+3+5+...+2n-3)$

$t_n=$sum of $(n-1)$ odd numbers

$={(n-1)}^2=n^2-2n+n$

Sum $=\sum t_n$

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

Hence, option (C) is the correct answer.