$2+4+7+11+16+\ldots$ to $n$ terms $=$

$\frac{1}{6}\left(n^2+3 n+8\right)$

$\frac{1}{6}\left(n^2-3 n+8\right)$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Medium

1 of 1

Application +4 | -1

2+4+7+11+16+\ldots

n

terms

=

View answer and explanation

Correct option: B

\frac{n}{6}\left(n^2+3 n+8\right)

Option B (

\frac{n}{6}(n^2+3n+8)

) is correct because it represents the sum of the series 2 + 4 + 7 + 11 + 16 + ... to n terms, derived by finding the nth term and summing.

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