Question

Show that the sum to $n$ terms of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+...$ is equal to $n-\frac{1}{2}(1-{\left(\frac{1}{3}\right)}^n)$.

Answer 1

The given series can be written as
$(1-\frac{1}{3})+(1-\frac{1}{9})+(1-\frac{1}{27})+(1-\frac{1}{81})+...$ upto $n$ terms
$=n-(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n})$
$=n-\frac{\frac{1}{3}(1-{\left(\frac{1}{3}\right)}^n)}{1-\frac{1}{3}}=n-\frac{1}{2}(1-{\left(\frac{1}{3}\right)}^n)$