JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q4

MathematicsSingle correct+4 / -1Question 4 of 90

The sum of the infinite series $1+\dfrac56+\dfrac{12}{6^2}+\dfrac{22}{6^3}+\dfrac{35}{6^4}+\dfrac{51}{6^5}+\dfrac{70}{6^6}+\dots$ is equal to:

\dfrac{425}{216}

\dfrac{429}{216}

\dfrac{288}{125}

\dfrac{280}{125}

Answer and worked solution

Official answer: C

Terms. Numerators $1,5,12,22,35,\dots$ are $\frac{n(3n-1)}2$ , so the sum is $\sum_{n\ge1}\frac{n(3n-1)}2x^{n-1}$ with $x=\frac16$ .

Series. $\sum n x^{n-1}=\frac1{(1-x)^2}$ , $\sum n^2x^{n-1}=\frac{1+x}{(1-x)^3}$ . At $x=\frac16$ : $\frac{36}{25}$ and $\frac{252}{125}$ .

Sum. $\frac12\left(3\cdot\frac{252}{125}-\frac{36}{25}\right)=\frac{288}{125}$ .

(C)

Topics:Arithmetico-geometric series Special series sums

JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 4