Question

The sum of the series
$\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+....$ upto 20 terms is

• A. $\frac{205}{3}$
• B. $\frac{200}{3}$
• C. $\frac{220}{3}$
• D. $\frac{210}{3}$

Answer 1

Answer: (C)

Let $S=\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+...$ upto 20 terms
Let $t_n$ be $n^{th}$ term of series then $t_n=\frac{1^2+2^2+3+....+n^2}{n\cdot (n+1)}$
$=\frac{\sum n^2}{n\left(n+1\right)}=\frac{n\left(n+1\right)\left(2n+1\right)}{6\cdot n\left(n+1\right)}=\frac{2n+1}{6}$
Taking summation on both sides
$\sum _{n=1}^{20}{t}_n=\frac{2}{6}\sum _{n=1}^{20}n+\frac{1}{6}\sum _{n=1}^{20}1$
$=\frac{1}{3}\times \left(\frac{20(20+1)}{2}\right)+\frac{1}{6}\times 20$
$=\frac{1}{3}\times \left(\frac{20\times 21}{2}\right)+\left(\frac{10)}{3}\right)$
$=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}$