JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q12

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MathematicsSingle correct+4 / -1Question 12 of 90

Let $\{a_n\}_{n=0}^\infty$ be a sequence such that $a_0=a_1=0$ and $a_{n+2}=2a_{n+1}-a_n+1$ for all $n\ge0$ . Then, $\displaystyle\sum_{n=2}^\infty\frac{a_n}{7^n}$ is equal to :

\dfrac6{343}

\dfrac7{216}

\dfrac8{343}

\dfrac{49}{216}

Answer and worked solution

Official answer: B

Sequence. Second differences are 1, so $a_n=\frac{n(n-1)}2$ .

Sum. $\frac12\sum n(n-1)x^n=\dfrac{x^2}{(1-x)^3}$ at $x=\frac17$ : $\dfrac{1/49}{216/343}=\dfrac7{216}$ .

(B)

Topics:Special series sums Arithmetico-geometric series

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 12