Find the sum of the special series $1^{2} + 2^{2} + 3^{2} + . . . . . . . . . + 35^{2}$ .

11910

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12910

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13910

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14910

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Solution

The correct option is D 14910
$\sum_{i = 1}^{n} n^{2} = \frac{n (n + 1) (2 n + 1)}{6} S_{n} = \frac{35 \times 36 \times 71}{6} = 14910$

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1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . .

Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …

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1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . .

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(1^{2} + 1) 1! + (2^{2} + 1) 2! + (3^{2} + 1) 3! + . . . . . . (n^{2} + 1) n!

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