JEE Main · 2009 · 12 Apr 2009 · Offline · Sequences and Series

Let $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\cdots$ and write down $\frac{1}{3}S$ shifted by one place. Subtracting, $S-\frac{1}{3}S$ leaves $1$ plus a series whose numerators settle to the constant difference 4, forming a geometric tail that sums neatly. Carrying out the arithmetic gives $\frac{2}{3}S=2$, so $S=3$.