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Strictly Increasing Functions

12+32+52+72+9...

Question

$1^{2} + 3^{2} + 5^{2} + 7^{2} + 9^{2}$ =___.

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Solution

Sum of squares of $1^{s t}$ 10 natural numbers - Sum of squares of $1^{s t}$ 5 even natural numbers

Sum of squares of $1^{s t}$ ‘n’ natural numbers = $\frac{n (n + 1) (2 n + 1)}{6}$

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Similar questions

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

3 n

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} + 3^{2} + 5^{2} + 7^{2} + \dots

n

(x - 1)^{2} + (y - 3)^{2} + (z - 5)^{2} + (t - 7)^{2} = 0

x y z t + 16

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