Question

Find the sum of the series $1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+...$ to n terms.

Answer 1

Let $S=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+...+\frac{3n-2}{5^{n-1}}$ (i)
$\left(\frac{1}{2}\right)S=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+...+\frac{3n-5}{5^{n-1}}+\frac{3n-2}{5^n}$ (ii)
(i)-(ii) $\Rightarrow$ $\frac{4}{5}S=1+\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+...+\frac{3}{5^{n-1}}-\frac{3n-2}{5^n}$
$\frac{4}{5}S=1+\frac{\frac{3}{5}(1-{\left(\frac{1}{5}\right)}^{n-1})}{1-\frac{1}{5}}-\frac{3n-2}{5^n}=1+\frac{3}{4}-\frac{3}{4}\times \frac{1}{5^{n-1}}-\frac{3n-2}{5^n}=\frac{7}{4}-\frac{12n+7}{{4.5}^n}$
$\therefore$ $S=\frac{35}{16}-\frac{(12n+7)}{{16.5}^{n-1}}$