Let $\left\{a_{n}\right\}_{n=0}^{\infty}$ be a sequence such that $\mathrm{a}_{0}=\mathrm{a}_{1}=0$ and $a_{n+2}=2 a_{n+1}-a_{n}+1$ for all $n \geq 0$. Then, $\sum_{n=2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Sequences and Series

Let

{a_{n}}_{n = 0}^{\infty}

be a sequence such that

a_{0} = a_{1} = 0

and

a_{n + 2} = 2 a_{n + 1} - a_{n} + 1

for all

n \geq 0

. Then,

\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}

is equal to

Answer: (B)

Solving the recurrence

a_{n + 2} = 2 a_{n + 1} - a_{n} + 1

with

a_{0} = a_{1} = 0

gives the triangular numbers

a_{n} = \frac{n ( n - 1 )}{2}

for

n \geq 2

(indeed

a_{2} = 1, a_{3} = 3, a_{4} = 6, \dots

). With

x = \frac{1}{7}

the required sum is

\frac{1}{2} n \geq 2 \sum n (n - 1) x^{n}

. Using

n \geq 2 \sum n (n - 1) x^{n} = \frac{2 x ^{2}}{( 1 - x ) ^{3}}

and substituting

x = \frac{1}{7}

gives

\frac{2/49}{( 6/7 ) ^{3}} = \frac{2}{49} \cdot \frac{343}{216} = \frac{14}{216}

, so the sum is

\frac{7}{216}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option