11^2 + 12^2 + 13^2 + dots + 20^2 = ? | vedclass

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Question

(C) The sum of squares of first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.We need to find the sum $S = 11^2 + 12^2

Vedclass · Accepted Answer

$2485$

Vedclass · Answer

(C) The sum of squares of first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.We need to find the sum $S = 11^2 + 12^2 + \dots + 20^2$.This can be expressed as $S = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{10} k^2$.For $n = 20$,$\sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 	imes 20 + 1)}{6} = \frac{20 	imes 21 	imes 41}{6} = 10 	imes 7 	imes 41 = 2870$.For $n = 10$,$\sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 	imes 10 + 1)}{6} = \frac{10 	imes 11 	imes 21}{6} = 5 	imes 11 	imes 7 = 385$.Therefore,$S = 2870 - 385 = 2485$.

$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

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