1-2-2-2-3-2-4-2-7-2 - nbsp-120140105

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Sum of Squares of n Natural Numbers

1^2+2^2+3^2+4...

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

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Similar questions

1^{2} = \frac{1}{6} [1 \times (1 + 1) \times (2 \times 1 + 1)] 1^{2} + 2^{2} = \frac{1}{6} [2 \times (2 + 1) \times (2 \times 2 + 1)] 1^{2} + 2^{2} + 3^{2} = \frac{1}{6} [3 \times (3 + 1) \times (2 \times 3 + 1)] 1^{2} + 2^{2} + 3^{2} + 4^{2} = \frac{1}{6} [4 \times (4 + 1) \times (2 \times 4 + 1)]

(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3}

(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}

[{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}] \div {(\frac{1}{4})}^{- 3}

(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2}

1^{2} + 2^{2} + 3^{2} + 4^{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} + 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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