JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Sequences and Series

Let $S={\sum }_{k=1}^{20}\frac{k}{2^k}$. This is an arithmetico-geometric sum, so multiply by $\frac{1}{2}$ and subtract. With $\frac{S}{2}={\sum }_{k=1}^{20}\frac{k}{2^{k+1}}$, lining the two series up term by term leaves a geometric series plus one stray end term: $\frac{S}{2}=\frac{1}{2}+\left(\frac{1}{2^2}+\cdots +\frac{1}{2^{20}}\right)-\frac{20}{2^{21}}$. Summing the geometric part, $\frac{S}{2}=1-\frac{1}{2^{20}}-\frac{20}{2^{21}}=1-\frac{11}{2^{20}}$. Doubling, $S=2-\frac{11}{2^{19}}$.