Question

The sum of$n$ terms of $(2n-1)+2(2n-3)+3(2n-5)+...$is

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

Answer 1

The correct option is D

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

Explanation for the correct option:

Find the required sum.

Given: series $(2n-1)+2(2n-3)+3(2n-5)+\dots n$ terms

Here $t_r=r\left[2n-(2r-1)\right]$

Sum,$S=\sum _{r=1}^n{t}_r$

$$\begin{gathered} =\sum _{r=1}^{n}r\left[2n-(2r-1)\right] \\ =2n\sum r–2\sum r^2+\sum r \\ =2n\times \frac{n(n+1)}{2}–2\times \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \\ =\frac{\left[6n^2\right(n+1)–2n(n+1\left)\right(2n+1)+3n(n+1\left)\right]}{6} \\ =\frac{n(n+1)[6n-4n-2+3]}{6} \\ =\frac{n(n+1)(2n+1)}{6} \end{gathered}$$

Therefore, the sum of$n$ terms of $(2n-1)+2(2n-3)+3(2n-5)+...$is $\frac{n(n+1)(2n+1)}{6}$.

Hence, option (D) is the correct answer.