Question

The sum of the series

$1 + 3 + 6 + 10 + - - - n \: terms$
is ______​

Answer 1

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

Given series is

$\rm \:S_{n} = 1 + 3 + 6 + 10 + - - - - \: n \: terms$

can be rewritten as

$\rm \: S_{n} = 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + - - - - \: n \: terms$

So, nth term of the series is

$\rm \: T_{n} = 1 + 2 + 3 + - - - + n$

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

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

Now,

$\rm \: S_{n} = T_{1} + T_{2} + T_{3} + - - - + T_{n}$

$\rm \: S_{n} = \displaystyle\sum_{k=1}^n\rm T_{k}$

$\rm \: S_{n} = \displaystyle\sum_{k=1}^n\rm \dfrac{1}{2} \bigg( {k}^{2} + k \bigg)$

$\rm \: S_{n} = \dfrac{1}{2} \displaystyle\sum_{k=1}^n\rm \bigg( {k}^{2} + k \bigg)$

$\rm \: S_{n} = \dfrac{1}{2}\bigg(\dfrac{n(n + 1)(2n + 1)}{2} + \dfrac{n(n + 1)}{2} \bigg)$

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

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

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

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

$\rm\implies \:\rm \: S_{n} = \dfrac{n(n + 1)(n + 2)}{6}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\sum_{k=1}^n\rm k \: = \: \dfrac{n(n + 1)}{2} }\\ \\ \bigstar \: \bf{\displaystyle\sum_{k=1}^n\rm {k}^{2} \: = \: \dfrac{n(n + 1)(2n + 1)}{6} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$