Question

Find the sum of the series

$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + - - - - n \: terms$
​

Answer 1

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

Given series is

$\rm \: \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + - - - - n \: terms$

can be further rewritten as

$\rm \: = \dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + - - - - n \: terms$

As 1, 2, 3, ... are in AP

So, nᵗʰ term = 1 + (n - 1)×1 = 1 + n - 1 = n

Also, 2, 3, 4, ... are in AP

So, nᵗʰ term = 2 + (n - 1)×1 = 2 + n - 1 = n + 1

So, above series can be rewritten as

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

can be further rewritten as

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

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

$\rm \: = 1 - \cancel \dfrac{1}{2} + \cancel \dfrac{1}{2} - \cancel \dfrac{1}{3} +\cancel \dfrac{1}{3} - \cancel \dfrac{1}{4} + - - +\cancel \dfrac{1}{n} - \dfrac{1}{n + 1}$

$\rm \: = 1 - \dfrac{1}{n + 1}$

$\rm \: = \dfrac{n + 1 - 1}{n + 1}$

$\rm \: = \dfrac{n}{n + 1}$

Hence,

$\boxed{\rm{\rm\implies \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + - - - n \: terms = \frac{n}{n + 1} \: }}$

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

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\sum_{k=1}^n\rm k \: = \: \dfrac{n(n + 1)}{2} }\\ \\ \bigstar \: \bf{\displaystyle\sum_{k=1}^n\rm {k}^{2} \: = \: \dfrac{n(n + 1)(2n + 1)}{6} }\\ \\ \bigstar \: \bf{\displaystyle\sum_{k=1}^n\rm {k}^{3} \: = \: \bigg(\dfrac{n(n + 1)}{2} \bigg)^{2} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

[Refer to the above attachment for your answer]

Hope you have satisfied. ⚘