Test Index

Sequence and Series

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Question : 39 of 86

Marks: +1, -0

Find the sum of

n

terms of the series

1+9+24

+46+75+\dots

$\;\frac{n(n+1)(7 n-4)}{6}$
$\;\frac{n(2 n+1)(4 n-3)}{6}$
$\;\frac{n(2 n+1)(2 n-3)}{6}$
None

Solution:

Concept:

The series follows a quadratic pattern because its second difference is constant (7).

Explanation:

The general term is

T_n = \frac{n}{2}(7n-5)

.

Sum of

n

terms:

S_n = \frac{1}{2} \left[ 7 \sum k^2 - 5 \sum k \right]

, with

\sum k = \frac{n(n+1)}{2}

and

\sum k^2 = \frac{n(n+1)(2n+1)}{6}

.

Simplifying:

S_n = \frac{1}{12} n(n+1)(14n-8) = \frac{n(n+1)(7n-4)}{6}

.

Answer:

\frac{n(n+1)(7n-4)}{6}

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