Question

Find the sum of the series

$C_0 + \frac{C_1}{2} + \frac{C_2}{3} + - - - + \frac{C_n}{n + 1}$

Please give proper explanation​

Answer 1

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

From Binomial expansion, we have

$\rm :\longmapsto\: {(1 + x)}^{n} = C_0 + C_1x + C_2 {x}^{2} + - - - + C_n {x}^{n}$

On integrating both sides w. r. t. x, between the limits x = 0 and x = 1, we get

$\rm :\longmapsto\: \displaystyle\int_0^1\rm {(1 + x)}^{n}dx = \displaystyle\int_0^1\rm \bigg[C_0 + C_1x + C_2 {x}^{2} + - - - + C_n {x}^{n}\bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm k \: f(x) \: dx \: = \: k\displaystyle\int \: f(x) \: dx \: }}}$

and

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c\: }}}$

So, using this we get

$\rm \: \bigg[\dfrac{ {(1 + x)}^{n + 1} }{n + 1} \bigg]_0^1 = C_0\bigg[\dfrac{x}{1} \bigg]_0^1 + C_1\bigg[\dfrac{ {x}^{2} }{2} \bigg]_0^1 + - - - + C_n\bigg[\dfrac{ {x}^{n + 1} }{n + 1} \bigg]_0^1$

$\rm \: \dfrac{ {2}^{n + 1} }{n + 1} - \dfrac{1}{n + 1} = C_0(1 - 0) + \dfrac{C_1}{2}(1 - 0) + \dfrac{C_2}{3}(1 - 0) + - + \dfrac{C_n}{n + 1}(1 - 0)$

$\rm \: \dfrac{ {2}^{n + 1} - 1}{n + 1} = C_0 + \dfrac{C_1}{2} + \dfrac{C_2}{3}+ - + \dfrac{C_n}{n + 1}$

Hence,

$\red{\boxed{\tt{ \rm \:C_0 + \dfrac{C_1}{2} + \dfrac{C_2}{3}+ - - - + \dfrac{C_n}{n + 1} = \dfrac{ {2}^{n + 1} - 1}{n + 1}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\boxed{\tt{ {(1 + x)}^{n} = C_0 + C_1x + C_2 {x}^{2} + - - - + C_n{x}^{n}}}$

$\boxed{\tt{ {(1 - x)}^{n} = C_0 - C_1x + C_2 {x}^{2} + - - - + {( - 1)}^{n} C_n{x}^{n}}}$

$\boxed{\tt{ C_0 + C_1 + C_2 + - - - + C_n = {2}^{n}}}$

$\boxed{\tt{ C_0 - C_1 + C_2 + - - - + {( - 1)}^{n} C_n = 0}}$

$\boxed{\tt{ C_0 + C_2 + C_4 + - - - = C_1 + C_3 + C_5 + - - - = {2}^{n - 1}}}$