The value of $100^{2} - 99^{2} + 98^{2} - 97^{2} . . . . . . . . . - 3^{2} + 2^{2} - 1^{2}$ is:

5050

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4950

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5000

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25000

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Solution

The correct option is A $5050$
Consider the given series.
$100^{2} - 99^{2} + 98^{2} - 97^{2} + . . . . . . . - 3^{2} + 2^{2} - 1^{2}$

Now,
$\Rightarrow (100^{2} - 99^{2}) + (98^{2} - 97^{2}) + . . . . . . . - 3^{2} + (2^{2} - 1^{2})$
$\Rightarrow (100 + 99) (100 - 99) + (98 + 97) (98 - 97) + . . . . . . . + (4 + 3) (4 - 3) + (2 + 1) (2 - 1)$
$\Rightarrow (199) + (195) + . . . . . . . + (7) + (3)$

Now, the no. of terms
$199 = 3 + (n - 1) 4$
$n = 50$

Therefore, the sum will be
$= \frac{50}{2} (3 + 199)$
$= 25 \times 202$
$= 5050$

Hence, this is the answer.

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Similar questions

$S = 100^{2} - 99^{2} + 98^{2} - 97^{2} + . . . . . . . . . . . . . . . . . . . . . . . . + 2^{2} - 1^{2};$ what is the value of S?

100^{2} - 99^{2} + 98^{2} - 97^{2} + 96^{2} + 95^{2} + . . . . = S

S = ?

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

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

(x + 100)^{2} + (x + 99)^{2} + \dots \dots + (x + 1)^{2} + x^{2} + (x - 1)^{2} + (x - 2)^{2} + \dots . + (x - 100)^{2}

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