JEE Main 2026 Apr 6 Shift 2, Mathematics Q6: Summation of Series

JEE Main2026Apr 6, Shift 2Mathematics

The sum $1 + \frac{1}{2} (1^{2} + 2^{2}) + \frac{1}{3} (1^{2} + 2^{2} + 3^{2}) + \dots$ upto 10 terms is equal to:

A
130
B
155
C
$\frac{315}{2}$
D
$\frac{325}{2}$

Solution

General term: $T_{n} = \frac{1}{n} \sum_{k = 1}^{n} k^{2} = \frac{1}{n} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{( n + 1 ) ( 2 n + 1 )}{6} = \frac{2 n ^{2} + 3 n + 1}{6}$ .

$S_{10} = \sum_{n = 1}^{10} T_{n} = \frac{1}{6} [2 \sum n^{2} + 3 \sum n + \sum 1]$ .

With $\sum_{1}^{10} n^{2} = \frac{10 \cdot 11 \cdot 21}{6} = 385$ and $\sum_{1}^{10} n = 55$ :

$S_{10} = \frac{1}{6} [2 (385) + 3 (55) + 10] = \frac{1}{6} [770 + 165 + 10] = \frac{945}{6} = \frac{315}{2}$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q6: Summation of Series

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