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JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Sequences and Series

The sum

\sum_{n = 1}^{7} \frac{n ( n + 1 ) ( 2 n + 1 )}{4}

is equal to

Answer: 504

Solution

Split the summand as

\frac{n ( n + 1 ) ( 2 n + 1 )}{4} = \frac{2 n ^{3} + 3 n ^{2} + n}{4}

, so the sum over

n = 1

to

7

is

\frac{1}{4} [2 \sum n^{3} + 3 \sum n^{2} + \sum n]

. Using

\sum n = \frac{7 \cdot 8}{2} = 28

,

\sum n^{2} = \frac{7 \cdot 8 \cdot 15}{6} = 140

and

\sum n^{3} = 2 8^{2} = 784

, this is

\frac{1}{4} [2 (784) + 3 (140) + 28] = \frac{1568 + 420 + 28}{4} = \frac{2016}{4} = 504

.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer

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