Question

${1.2}^2+{2.3}^2+{3.4}^2+...$ to $n$ terms.

• A. $\frac{1}{12}n(n+1)(3n^2+11n+10)$
• B. $\frac{1}{6}n(n-1)(3n^2+7n+5)$
• C. $\frac{1}{3}n(n+1)(3n^2-7n+5)$
• D. $\frac{1}{2}n(n+1)(3n^2+7n+5)$

Answer 1

The correct option is A $\frac{1}{12}n(n+1)(3n^2+11n+10)$

General term $T_r=r(2+(r-1)1{)}^2=r(r+1{)}^2=r^3+2r^2+r$
$\therefore$ Sum $=\sum _{r=1}^n{T}_r=\sum _{r=1}^n{r}^3+\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r$
$=(\frac{n(n+1)}{2}{)}^2+\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{1}{12}n(n+1)(3n^2+11n+10)$