JEE Advanced 2021 Paper 1 · Mathematics · Sequences and Series · medium

A sequence is defined by a1=1 and an+1=(an)/(1+n an) for n≥1. The value of Σn=1^101/an is

Single correct option

A sequence is defined by

a_1=1

and

a_{n+1}=\dfrac{a_n}{1+n\,a_n}

for

n\ge1

. The value of

\displaystyle\sum_{n=1}^{10}\dfrac{1}{a_n}

Show answer and solution

Answer: (A) $175$

Take reciprocals:

\frac{1}{a_{n+1}}=\frac{1+n a_n}{a_n}=\frac{1}{a_n}+n

. With

b_n=1/a_n

we get

b_{n+1}-b_n=n

and

b_1=1

, so

b_n=1+\sum_{j=1}^{n-1}j=1+\frac{n(n-1)}{2}

. Then

\sum_{n=1}^{10}b_n=10+\sum_{n=1}^{10}\binom n2=10+\binom{11}{3}=10+165=175

. A direct recursion for the first ten terms (

b_n=1,2,4,7,11,16,22,29,37,46

) confirms

175

. Option (B) forgets the

+1

in each term; (C) and (D) come from mis-evaluating

\sum\binom n2

or starting the sum at

n=0

Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake