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Sum of Squares of First N Natural Numbers

Which of the ...

Question

Which of the following expression gives the value of "s". "s" is given as $1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . . . . . . . . . . . . . + 7^{2}$

$s = \frac{7 \times (7 + 1) (2 \times 7 + 1)}{8}$

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$s = \frac{7 \times (7 + 1) (3 \times 7 + 1)}{6}$

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$s = \frac{7 \times (7 + 2) (2 \times 7 + 1)}{6}$

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$s = \frac{7 \times (7 + 1) (2 \times 7 + 1)}{6}$

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Solution

The correct option is D
$s = \frac{7 \times (7 + 1) (2 \times 7 + 1)}{6}$

The formula for the sum of the squares of first "n" natural numbers is:

$s = \frac{n \times (n + 1) \times (2 n + 1)}{6}$ . Here, n= 7

So, $s = \frac{7 \times (7 + 1) (2 \times 7 + 1)}{6}$

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Similar questions

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} + . . . . . . = . . . . . . . .

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} +

=

S = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . . . . . . . . . .

\infty

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + . . . . . . . . . . . . .

\infty

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . + 7^{2} =

12 - 7 - 3 ?

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