Question

The sum of the series $1+(1+2)+(1+2+3)+(1+2+3+4)+\dots \dots ..+(1+2+3+4+\dots +20)$ is

• A. 1470
• B. 1540
• C. 1610
• D. 1370

Answer 1

The correct option is B

1540

Find the sum of series:

Let the series,

$\begin{array}{rcl}S& =& 1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots +(1+2+3+4+\dots +20)\\ & =& T_1+T_2+T_3+T_4+\cdots +T_{20}......\left(1\right)\end{array}$

The term $T_r$ can be write it as follows.

$\begin{array}{rcl}{T}_r& =& 1+2+3\dots +r\\ & =& \frac{r(r+1)}{2}\\ & =& \frac{r^2+r}{2}\end{array}$

Formulae:

It is known that, sum of natural numbers and sum of the square of natural numbers are

$\begin{array}{rcl}\underset{r=1}{\overset{n}{\sum r}}& =& \frac{n(n+1)}{2}\end{array}$

$\begin{array}{rcl}\sum _{r=1}^n{r}^2& =& \frac{n(n+1)\left(2n+1\right)}{6}\end{array}$

Rewrite the equation $\left(1\right)$ by using the above-mentioned formulae.

$\begin{array}{rcl}S& =& \underset{r=1}{\overset{20}{\sum T_r}}\\ & =& \sum _{r=1}^{20}\frac{r^2+r}{2}\\ & =& \frac{1}{2}\left(\sum _{r=1}^{20}{r}^2+\sum _{r=1}^{20}r\right)\\ & =& \frac{1}{2}\left(\frac{\left(20\right)\left(20+1\right)\left(2\left(20\right)+1\right)}{6}+\frac{20\left(21\right)}{2}\right)\\ & =& \frac{1}{4}\left(\frac{20\left(21\right)\left(41\right)}{3}+20\left(21\right)\right)\\ & =& 1540\end{array}$

Hence, the option (b) is the correct answer.