Question

the sum of 1 + 2/5 + 3/25 + 4/125 + ...... upto nterms is​

Answer 1

Given :

The series as 1 + $\dfrac{2}{5 } + \dfrac{3}{25} + \dfrac{4}{125}$ +  ............. up to n terms

To Find :

The sum of  1 + $\dfrac{2}{5 } + \dfrac{3}{25} + \dfrac{4}{125}$ +  ............. up to n terms

Solution :

S = 1 + $\dfrac{2}{5 } + \dfrac{3}{25} + \dfrac{4}{125}$ +  ............. up to n terms           ............1

$\dfrac{S}{5}$  = $\dfrac{1}{5}$ +  $\dfrac{2}{25 } + \dfrac{3}{125} + \dfrac{4}{625}$ +  ............. up to n terms            ...........2

Subtracting eq 2 from eq 1

i.e  S - $\dfrac{S}{5}$  = [  1 + $\dfrac{2}{5 } + \dfrac{3}{25} + \dfrac{4}{125}$ +  ............. up to n terms ] - [  $\dfrac{1}{5}$ +  $\dfrac{2}{25 } + \dfrac{3}{125} + \dfrac{4}{625}$ +  ............. up to n terms ]

Or,   $\dfrac{4S}{5}$  = 1 +  [ ($\dfrac{2}{5}$  - $\dfrac{1}{5}$ ) + ( $\dfrac{3}{25}$  - $\dfrac{2}{25}$ ) + ( $\dfrac{4}{125}$ - $\dfrac{3}{125}$ ) +  ........... ]

Or,  $\dfrac{4S}{5}$  = 1 + [ $\dfrac{1}{5} + \dfrac{1}{25}+\dfrac{1}{125}$  + ............ ]

Now, $\dfrac{1}{5} + \dfrac{1}{25}+\dfrac{1}{125}$  + ............  is in Geometric progression sum

So,  $\dfrac{4S}{5}$  = 1 + [ $\dfrac{\dfrac{1}{5} }{1-\dfrac{1}{5} }$ ]

Or,  $\dfrac{4S}{5}$  = 1 + ( $\dfrac{\dfrac{1}{5} }{\dfrac{4}{5} }$ )

Or,  $\dfrac{4S}{5}$  = 1 + $\dfrac{1}{4}$

Or,  $\dfrac{4S}{5}$  =  $\dfrac{5}{4}$

∴      S = $\dfrac{25}{16}$

Hence, The sum of given series is $\dfrac{25}{16}$  .  Answer