JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
If $S=\frac{7}{5}+\frac{9}{5^{2}}+\frac{13}{5^{3}}+\frac{19}{5^{4}}+\ldots$, then $160S$ is equal to $\_\_\_\_\_\_$ .
Sequence and Series · Mixed Progressions (AP-GP)
Answer$305$
Solution
Formulas used. (1) $\sum_{k\ge1}x^k=\frac{x}{1-x}$, $\ \sum_{k\ge1}kx^k=\frac{x}{(1-x)^2}$, $\ \sum_{k\ge1}k^2x^k=\frac{x(1+x)}{(1-x)^3}$Step 1 - Numerators. $7,9,13,19$ have differences $2,4,6$, so they are $k^2-k+7$ (check $k=4$: $16-4+7=19$). Step 2 - The three sums with $x=\tfrac15$: $$\sum x^k=\frac14,\qquad \sum kx^k=\frac{1/5}{16/25}=\frac5{16},\qquad \sum k^2x^k=\frac{\frac15\cdot\frac65}{64/125}=\frac{15}{32}.$$ Step 3 - Combine. $$S=\frac{15}{32}-\frac5{16}+7\cdot\frac14=\frac{15-10+56}{32}=\frac{61}{32}.$$ Step 4. $$160S=5\times61=305 .$$ Answer: 305