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Composite Function

11 2 + 12 2 +...

Question

$(11^{2} + 12^{2} + 13^{2} + . . . . . 20^{2}) =$ ?

A

385

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B

2485

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C

2870

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D

3255

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Solution

The correct option is B $2485$
(112+122+132+.....202)=(12+22+32+.....202)−(12+22+32+.....102)[Ref:(12+22+32+.....n2)=16n(n+1)(2n+1)]

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Similar questions

(11)^{2} + (12)^{2} + (13)^{2} + \dots + (20)^{2}

10^{2} + 11^{2} + 12^{2} + . . . . + 20^{2}

Q. Sum infinite terms of the series

{cot}^{- 1} (1^{2} + \frac{3}{4}) + {cot}^{- 1} (2^{2} + \frac{3}{4}) + {cot}^{- 1} (3^{2} + \frac{3}{4}) + \dots

Q. Sum to infinite terms of the series

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Q. The sum of the infinite terms of the series

{cot}^{- 1} (1^{2} + \frac{3}{4}) + {cot}^{- 1} (2^{2} + \frac{3}{4}) + {cot}^{- 1} (3^{2} + \frac{3}{4}) + \dots

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