JEE Main 2022 (Online) 29th June Evening ShiftQuestion 1
The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots$ is equal to:
Sequence and Series · Mixed Progressions (AP-GP)
Solution
Formulas used. (1) $\sum_{n\ge1}n\,t^{n-1}=\frac1{(1-t)^2}$(2) $\sum_{n\ge1}n^2t^{n-1}=\frac{1+t}{(1-t)^3}$Step 1 - Numerators. $1,5,12,22,35,51,70$ have differences $4,7,10,13,\dots$ (second difference $3$). They are $\dfrac{n(3n-1)}{2}$: check $n=3$: $\dfrac{3\cdot8}{2}=12$. Step 2 - Write the series. With $t=\dfrac16$: $$S=\sum_{n\ge1}\frac{n(3n-1)}{2}\,t^{n-1}=\frac12\left[3\sum n^2t^{n-1}-\sum n\,t^{n-1}\right].$$ Step 3 - The two sums. $$\sum n^2t^{n-1}=\frac{7/6}{(5/6)^3}=\frac76\cdot\frac{216}{125}=\frac{252}{125},\qquad \sum n\,t^{n-1}=\frac1{(5/6)^2}=\frac{36}{25}=\frac{180}{125}.$$ Step 4. $$S=\frac12\left[\frac{756}{125}-\frac{180}{125}\right]=\frac12\cdot\frac{576}{125}=\frac{288}{125}.$$ Answer: (C)