JEE Main 2024 · Mathematics · Sequences and Series · medium

The sum of the first 20 terms of the series 3+7+13+21+31+… is

Numerical answer

The sum of the first

20

terms of the series

3+7+13+21+31+\cdots

Show answer and solution

Answer: 3100

The successive differences

4,6,8,10,\ldots

form an arithmetic progression, so the general term is quadratic. Writing

T_n=an^2+bn+c

and using

T_1=3

T_2=7

T_3=13

gives

a=1

b=1

c=1

, i.e.

T_n=n^2+n+1

(check:

T_4=21

T_5=31

). Then

\sum_{n=1}^{20}T_n=\dfrac{20\cdot21\cdot41}{6}+\dfrac{20\cdot21}{2}+20=2870+210+20=3100

. As a check,

T_4=16+4+1=21

and

T_5=25+5+1=31

, matching the given terms, and the total of the first four terms is

3+7+13+21=44

, equal to

30+10+4

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