The value of $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . + 11^{2}$ is equal to

$55$

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$66$

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$77$

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$88$

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Solution

The correct option is B
$66$

The explanation for the correct answer.
Given: $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . + 11^{2}$
The above expression can be written as
$= 1^{2} - 2^{2} + 2^{2} - 2^{2} + 3^{2} - 4^{2} + 4^{2} - 4^{2} + . . . . + 11^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + 11^{2} - 2 (2^{2} + 4^{2} + 6^{2} + 8^{2} + 10^{2})$
Now take $2^{2}$ common form second term
$= 1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + 11^{2} - 2^{3} (1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2})$
Now use the formula for the sum of squares of natural numbers $\frac{n (n + 1) (2 n + 1)}{6}$
$= \frac{11 (11 + 1) (2 \times 11 + 1)}{6} - 8 (\frac{5 (5 + 1) (2 \times 5 + 1)}{6}) = \frac{11 \times 12 \times 23}{6} - 8 \times \frac{5 \times 6 \times 11}{6} = 506 - 440 = 66$
Hence $Option (B)$ is the correct answer.

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