Question

Find the sum of the series,
(1) + (1+2) + (1+2+3) + ... to nth term.​

Answer 1

Given series :-

$\small \longrightarrow (1) + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + n)$

Here, each term is a sum of n natural number. Therefore, for very general term of the series, we have a formula :

• $\boxed{ \tt{T_r= \dfrac{n(n+1)}{2}}}$

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Now, the sum of the series is the sum of all terms. That means,

$\displaystyle \rm Sum\:of\:series = \sum\limits_{r=1}^n T_r$

$\displaystyle \rm Sum\:of\:series = \sum\limits_{r=1}^n \dfrac{r(r + 1)}{2}$

$\displaystyle \rm Sum\:of\:series = \sum\limits_{r=1}^n \dfrac{ {r}^{2} + r}{2}$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[\sum\limits_{r=1}^n (r^2 +r) \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[\sum\limits_{r=1}^n (r^2) +\sum\limits_{r=1}^n (r) \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n + 1)}{2}\right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)(2n + 1)+3n(n+1)}{6} \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)[(2n + 1)+3]}{6} \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)[2n + 1+3]}{6} \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)[2n + 4]}{6} \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)2[n + 2]}{6} \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)\not2[n + 2]}{\not6} \right]$

$\displaystyle \rm Sum\:of\:series = \dfrac{1}2 \left[ \dfrac{n(n + 1)(n + 2)}{3} \right]$

$\displaystyle \rm Sum\:of\:series = \left[ \dfrac{n(n + 1)(n + 2)}{6} \right]$

Therefore, the required sum of series is,

• $\boxed{ \mathfrak{\dfrac{n(n + 1)(n + 2)}{6}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\:1 + (1 + 2) + (1 + 2 + 3) + - - - n \: terms$

Let first find the nth term of the series.

Since,

$\rm :\longmapsto\:a_1 = 1$

$\rm :\longmapsto\:a_2 = 1 + 2$

$\rm :\longmapsto\:a_3 = 1 + 2 + 3$

So,

$\rm \implies\:a_n = 1 + 2 + 3 + - - + n$

$\rm \implies\:a_n = \displaystyle\sum_{n=1}^n \: n$

$\red{\rm \implies\:a_n = \dfrac{n(n + 1)}{2} = \dfrac{ {n}^{2} + n}{2} }$

So, Sum of the given series is given by

$\rm :\longmapsto\:S_n = \displaystyle\sum_{n=1}^na_n$

$\rm :\longmapsto\:S_n = \displaystyle\sum_{n=1}^n \frac{ {n}^{2} + n}{2}$

$\rm :\longmapsto\:S_n =\dfrac{1}{2} \displaystyle\sum_{n=1}^n ( {n}^{2} + n)$

$\rm :\longmapsto\:S_n = \dfrac{1}{2}\bigg(\dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n + 1)}{2} \bigg)$

$\rm :\longmapsto\:S_n = \dfrac{n(n + 1)}{4}\bigg(\dfrac{2n + 1}{3} + 1\bigg)$

$\rm :\longmapsto\:S_n = \dfrac{n(n + 1)}{4}\bigg(\dfrac{2n + 1 + 3}{3}\bigg)$

$\rm :\longmapsto\:S_n = \dfrac{n(n + 1)}{4}\bigg(\dfrac{2n + 4}{3}\bigg)$

$\rm :\longmapsto\:S_n = \dfrac{n(n + 1)}{4}\bigg(\dfrac{2(n + 2)}{3}\bigg)$

$\rm \implies\:\boxed{ \tt{ \: S_n = \frac{n(n + 1)(n + 2)}{6} \: }}$

Result Used :-

$\boxed{ \tt{ \: \displaystyle\sum_{n=1}^n \: n \: = \: \frac{n(n + 1)}{2} \: }}$

$\boxed{ \tt{ \: \displaystyle\sum_{n=1}^n \: {n}^{2} \: = \: \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\boxed{ \tt{ \: \displaystyle\sum_{n=1}^n {n}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2} \: }}$