JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q5

MathematicsSingle correct+4 / -1Question 5 of 90

Let $S=2+\dfrac67+\dfrac{12}{7^2}+\dfrac{20}{7^3}+\dfrac{30}{7^4}+\dots$ . Then $4S$ is equal to

\left(\dfrac73\right)^2

\dfrac{7^3}{3^2}

\left(\dfrac73\right)^3

\dfrac{7^2}{3^3}

Answer and worked solution

Official answer: C

Series. $S=\displaystyle\sum_{n\ge0}(n+1)(n+2)x^n=\frac2{(1-x)^3}$ with $x=\frac17$ : $S=2\left(\frac76\right)^3$ .

Value. $4S=\dfrac{8\cdot343}{216}=\left(\dfrac73\right)^3$ .

(C)

Topics:Special series sums Binomial expansion

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 5