Question

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

Answer 1

Let $S=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+....+\frac{3n-2}{5^{n-1}}$;
$\therefore \frac{1}{5}S=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+....+\frac{3n-5}{5^{n-1}}+\frac{3n-2}{5^n}$;

Subtracting both equations, we get
$\therefore \frac{4}{5}S=1+(\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+....+\frac{3}{5^{n-1}})-\frac{3n-2}{5^n}$
$=1+\frac{3}{5}.\frac{1-\frac{1}{5^{n-1}}}{1-\frac{1}{5}}-\frac{3n-2}{5^n}$
$=1+\frac{3}{4}(1-\frac{1}{5^{n-1}})-\frac{3n-2}{5^n}$
$=\frac{7}{4}-\frac{12n+7}{4.5^n}$;
$\therefore S=\frac{35}{16}-\frac{12n+7}{{16.5}^{n-1}}$.