The sum 11^{2} + 12^{2} + cdots + 20^{2} + 21 | vedclass

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Question

(A) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$.To find the sum $

Vedclass · Accepted Answer

$2926$

Vedclass · Answer

(A) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$.To find the sum $11^{2} + 12^{2} + \cdots + 21^{2}$,we can express it as the difference between the sum of squares up to $21$ and the sum of squares up to $10$:Sum $= \sum_{k=1}^{21} k^{2} - \sum_{k=1}^{10} k^{2}$.For $n = 21$: $\frac{21(21+1)(2 	imes 21 + 1)}{6} = \frac{21 	imes 22 	imes 43}{6} = 7 	imes 11 	imes 43 = 3311$.For $n = 10$: $\frac{10(10+1)(2 	imes 10 + 1)}{6} = \frac{10 	imes 11 	imes 21}{6} = 5 	imes 11 	imes 7 = 385$.Therefore,the required sum is $3311 - 385 = 2926$.

The sum $11^{2} + 12^{2} + \cdots + 20^{2} + 21^{2} = ?$

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