$1^{2} + 2^{2} - 3^{2} + 4^{2} + 5^{2} - 6^{2} + 7^{2} + 8^{2} - 9^{2} + . . . .$ to $3 n$ terms.

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Solution

$1^{2} + 2^{2} - 3^{2} + 4^{2} + 5^{2} - 6^{2} + . . . . . . .$ to $3 n$ terms
$= (1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + . . . . . . . . + (3 n)^{2})$
$- 2 (3^{2} + 6^{2} + . . . . . . + (3 n)^{2})$
$= (1^{1} + 2^{2} + . . . . . . . . + (3 n)^{2}) - 2.9 (1^{2} + 2^{2} + . . . . . . . . + n^{2})$
$= \frac{3 n . (3 n + 1) (6 n + 1)}{6} - 18 \frac{n (n + 1) (2 n + 1)}{6}$
$= \frac{n (3 n + 1) (6 n + 1)}{2} - 3 n (n + 1) (2 n + 1)$
$= n [\frac{6 n^{2} - 9 n - 5}{2}]$

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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} + . . . . . . . . .

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)]

n

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

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