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

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Solution

The given series is $5^{2} + 6^{2} + 7^{2} + . . . + 20^{2}$
nth term , $a_{n} = {(n + 4)}^{2} = n^{2} + 8 n + 16$
$∴ S_{n} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} (k^{2} + 8 k + 16) = \sum_{k = 1}^{n} k^{2} + 8 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 16 = \frac{n (n + 1) (2 n + 1)}{6} + \frac{8 n (n + 1)}{2} + 16 n$
16th term is $(16 + 4)^{2} = 20^{2}$
$∴ S_{16} = \frac{16 (16 + 1) (2 \times 16 + 1)}{6} + \frac{8 \times 16 \times (16 + 1)}{2} + 16 \times 16 = \frac{16 (17) (33)}{6} + \frac{(8) \times 16 \times (16 + 1)}{2} + 16 \times 16 = \frac{16 (17) (33)}{6} + \frac{(8) (16) (17)}{2} + 256 = 1496 + 1088 + 256 = 2840 ∴ 5^{2} + 6^{2} + 7^{2} + . . . + 20^{2} = 2840$

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Similar questions

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

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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} +

=

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