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× ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Properties of Square Numbers

Solve 52+62+...

Question

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

Open in App

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

Suggest Corrections

Similar questions

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

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = . . .^{2}

Find the sum to n terms

$5^{2} + 6^{2} + 7^{2} + \dots + 20^{2}$

2^{2} + 4^{2} + 6^{2} + . . . + 20^{2}

1^{2} + 3^{2} + 5^{2} + . . . + 19^{2}

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} +

= 21^{2}

5^{2} +

+ 30^{2} = 31^{2}

6^{2} + 7^{2} +

=

Join BYJU'S Learning Program