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Sum of Squares of First N Natural Numbers

12+22+32+42+…...

Question

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

120

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140

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105

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Solution

The correct option is B 140
The sum of the squares of the first $n$ natural numbers is given by
$S_{n} = 1^{2} + 2^{2} + 3^{2} + . . . + n^{2}$
$= \frac{n (n + 1) (2 n + 1)}{6} .$

Here, $n = 7.$
$∴ S_{7} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . + 7^{2}$
$= \frac{7 (7 + 1) (2 \times 7 + 1)}{6}$
$= \frac{7 \times 8 \times 15}{6}$
$= 140$
Thus, $1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . + 7^{2} = 140.$

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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} +

=

1^{2} + 2^{2} - 3^{2} + 4^{2} + 5^{2} - 6^{2} + 7^{2} + 8^{2} - 9^{2} + . . . .

3 n

\frac{3}{1^{2} \times 2^{2}} - \frac{5}{2^{2} \times 3^{2}} + \frac{7}{3^{2} \times 4^{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

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