Question

The sum of the series $3·6+4·7+5·8+.......$ upto $\left(n-2\right)$ terms

• A. $n^3+n^2+n+2$
• B. $\frac{1}{6}\left(2n^3+12n^2+10n-84\right)$
• C. $n^3+n^2+n$
• D. none of these

Answer 1

The correct option is B

$\frac{1}{6}\left(2n^3+12n^2+10n-84\right)$

Explanation for the correct option

Given: $3·6+4·7+5·8+.......$ upto $\left(n-2\right)$ terms

Let, $S=3·6+4·7+5·8+.......$

last term $T_n=(r+2)(r+5)$

$=r^2+7r+10$

$$\begin{gathered} \therefore S=\sum _1^{n-2}\left(T_n\right) \\ \Rightarrow S=\sum _{r=1}^{n-2}\left(r^2+7r+10\right) \\ \Rightarrow S=\sum _{r=1}^{n-2}\left(r^2\right)+7\sum _{r=1}^{n-2}\left(r\right)+10\sum _{r=1}^{n-2}\left(1\right) \\ \Rightarrow S=\frac{\left(n-2\right)(n-2+1)(2\left(n-2\right)+1)}{6}+7\frac{\left(n-2\right)(\left(n-2\right)+1)}{2}+10\left(n-2\right) \\ \Rightarrow S=\frac{\left(n-2\right)(n-1)}{2}\left[\frac{2n-3}{3}+7\right]+10n-20 \\ \Rightarrow S=\frac{\left(n-2\right)(n-1)}{6}(2n-3+21)+10n-20 \\ \Rightarrow S=\frac{\left(n-2\right)(n-1)}{6}(2n+18)+10\left(n-2\right) \\ \Rightarrow S=\left(n-2\right)\left[\frac{(n-1)(n+9)}{3}+10\right] \\ \Rightarrow S=(n-2)\left[\frac{n^2+8n-9+30}{3}\right] \\ \Rightarrow S=\left(\frac{n-2}{3}\right)\left(n^2+8n+21\right) \\ \Rightarrow S=\frac{n^3-2n^2+8n^2+21n-16n-42}{3} \\ \Rightarrow S=\frac{n^3+6n^2+5n-42}{3} \\ \Rightarrow S=\frac{1}{6}\left(2n^3+12n^2+10n-84\right) \end{gathered}$$

Hence,$\mathrm{Option}\left(\mathrm{B}\right)$ is correct.