The sum to infinite terms of the series 1+23+63^2+103^3+143^4+ is:

Question

The sum to infinite terms of the series $1+\dfrac23+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\dfrac{14}{3^4}+\cdots$ is:

Accepted Answer

Let $S=1+\dfrac23+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\dfrac{14}{3^4}+\cdots$. Then $\dfrac13S=\dfrac13+\dfrac{2}{3^2}+\dfrac{6}{3^3}+\dfrac{10}{3^4}+\cdots$

Subtracting:
$$\frac23S=1+\frac13+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\cdots=\frac43+\frac{4}{3^2}+\frac{4}{3^3}+\cdots=\frac{4/3}{1-1/3}=2.$$

$S=3.$

Solutions · 1