Question

The sum to n terms of the series $3^2+8^2+{13}^2$ +.......

• A. $\frac{n(50n^2+15n-11)}{6}$
• B. $\frac{(50n^2+15n-11)}{6}$
• C. $\frac{n(50n^2+15n+11)}{6}$
• D. $\frac{n(50n^2-15n-11)}{6}$

Answer 1

The correct option is A $\frac{n(50n^2+15n-11)}{6}$

The sum to n terms of series $3^2+8^2+{13}^2+....$

The generate terms for above series can be written as:

$\mu r=(5r-2{)}^2$ where r=1 to n

$\therefore$ Sum up to n terms can be written as

$S_n={\sum }_{r=1}^n(5r-2{)}^2={\sum }_{r=1}^n(25r^2-20r+4)$

$=25{\sum }_{r=1}^n{r}^2-20{\sum }_{r=1}^{n}r+4{\sum }_{r=1}^n\left(1\right)$

$=25\frac{n(n+1)(2n+1)}{6}-20\frac{n(n+1)}{2}+4n$

$=\frac{25\left[\right(n^2+n\left)\right(2n+1\left)\right]-60(n^2+n)+24n}{6}$

$=\frac{25[2n^3+n^2+2n^2+n]-60n^2-60n+24n}{6}$

$=\frac{50n^3+75n^2+25n^2+35n-60n^2-60n+24n}{6}$

$=\frac{50n^3+14n^2-11n}{6}$

$Sn=\frac{n(50n^2+15n-11)}{6}$