The least value of ${(x + 100)}^{2} + {(x + 99)}^{2} + . . . . + {(x + 1)}^{2} + x^{2} + {(x - 1)}^{2} + {(x - 2)}^{2} + . . . . . . . + {(x - 100)}^{2}$ is

6767

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67670

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676700

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767600

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Solution

The correct option is C $676700$
Given

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

The least value of this series would at $x = 0$

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

$\Rightarrow (100)^{2} + (99)^{2} + . . . \dots + (1)^{2} + (1)^{2} + (2)^{2} + . . . \dots + (100)^{2}$

$\Rightarrow$ $2 [(1)^{2} + (2)^{2} + . . . \dots . + (100)^{2}]$

We know sum of square of $1$ st n natural number is given as $s = \frac{n (n + 1) (2 n + 1)}{6}$ where n is last term.

$\Rightarrow 2 [\frac{100 (100 + 1) (200 + 1)}{6}]$

$\Rightarrow \frac{100 \times 101 \times 201}{3}$

$= \frac{10100 \times 201}{3}$

$= 676700$ .

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Similar questions

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

100^{2} - 99^{2} + 98^{2} - 97^{2} . . . . . . . . . - 3^{2} + 2^{2} - 1^{2}

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} =

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

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