The sum of the series $1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . - 100^{2} ?$

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Solution

$1^{2} - 3^{2} + 5^{2} - 7^{2} + 9^{2} . . . - 99^{2} + - 101^{2}$
We can write this as
$y = 1^{2} + 5^{2} + 9^{2} . . {.101}^{2} - 3^{2} - 7^{2} - 11^{2} . . . - 99^{2} y = 1^{2} + (5^{2} - 3^{2}) + (9^{2} - 7^{2}) + . . {.101}^{2} - 99^{2}) a^{2} - b^{2} = (a + b) (a - b) y = 1 + (5 + 3) (5 - 3) + (9 + 7) (9 - 7) + . . . (101 + 99) (101 - 99) = 1 + 2 (3 + 5) + 2 (7 + 9) + 2 (11 + 13) . . . + 2 (99 + 101) y = 1 + 2 (3 + 5 + 7 + 11 + 13 + 15...99 + 101)$
These sequence 3, 5, 7 ...101 is an A.P with a = 3, d = 2
The sum of this sequence is $\frac{n}{2} (a + l)$
$a_{n} = a + n d$
$101 = 3 + n + 2$
$n^{2} = 49$
So, $S_{n} = \frac{49}{2} (3 + 101) = \frac{49}{2} \times 104$
$= 2548$
Hence $y = 1 + 2 (2548)$
$= 1 + 5096$
$= 5097$ .

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1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . + 99^{2} - 100^{2}

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