Question

What is the sum of the series $(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3})+(\frac{1}{5^4}+\frac{2}{5^5}+\frac{3}{5^6})+........$ to $7$ terms?

Answer 1

Let $S=(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3})+(\frac{1}{5^5}+\frac{2}{5^5}+\frac{3}{5^6})+.........$

$=(\frac{1}{5}+\frac{1}{5^4}+\frac{1}{5^7}+.......)+(\frac{2}{5^5}+\frac{2}{5^5}+\frac{2}{5^8}+....)+(\frac{3}{5^3}+\frac{3}{5^6}+\frac{3}{5^9}+.......)$

$=\frac{1}{5}(1+\frac{1}{5^3}+\frac{1}{5^6}+......)+\frac{2}{5^2}(1+\frac{1}{5^3}+\frac{1}{5^6}+.......)+\frac{3}{5^3}(1+\frac{1}{5^3}+\frac{1}{5^6}+.......)$

$=(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3})(1+\frac{1}{5^3}+\frac{1}{5^6}+.......)$

$=\frac{25+10+3}{125}\times \frac{1(1-\frac{1}{5^{3n}})}{1-\frac{1}{5^3}}$ [Sum of $G.P$ ]

$=\frac{38}{125}\times \frac{125}{124}(1-\frac{1}{5^{3n}})$

Putting $n=7,5=\frac{38}{124}(1-\frac{1}{5^{21}})\approx \mathrm{0.3064.}$