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
$\displaystyle \sum_{n =1}^{20} t_n = \frac{2}{6} \displaystyle \sum_{n =1}^{20}n+ \frac{1}{6} \displaystyle\sum_{n =1}^{20} 1$
$ = \frac{1}{3} \times \left(\frac{20(20 +1)}{2} \right) + \frac{1}{6} \times 20$
$ = \frac{1}{3} \times \left(\frac{20\times21}{2} \right)+ \left(\frac{10)}{3} \right) $
$ = \frac{210}{3} + \frac{10}{3} = \frac{220}{3}$