A]5^2+6^2+7^2+...+20^2 =(1^2+2^2+...+20^2) (1^2+2^2+3^2+4^2) sum of square #160; is n(n+1)(2n+1)/6 for n = 20 Sum=20(20+1)(40+1)/6=20× 21× ...

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Properties of Square Numbers

Solve 52+62+...

Question

Solve
$5^{2} + 6^{2} + 7^{2} + . . . + 20^{2}$

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Solution

$A] 5^{2} + 6^{2} + 7^{2} + . . . + 20^{2}$
$= (1^{2} + 2^{2} + . . . + 20^{2}) - (1^{2} + 2^{2} + 3^{2} + 4^{2})$
sum of square is $\frac{n (n + 1) (2 n + 1)}{6}$
for n = 20
$S u m = \frac{20 (20 + 1) (40 + 1)}{6} = \frac{20 \times 21 \times 41}{6} = 2870$
for n = 4
$s u m = \frac{4 (4 + 1) (8 + 1)}{6} = 2 \times 5 \times 3 = 30$
$∴ 5^{2} + 6^{2} + 7^{2} + . . {.20}^{2}$
= 2870 - 30
= 2840
$∴$ Required sum is 2840

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