If $1^{2} + 2^{2} + 3^{2} + . . . . . . . + 2003^{2} = (2003) (4007) (334)$ and
$(1) (2003) + (2) (2002) + (3) (2001) + . . . . . + (2003) (1) = (2003) (334) (x)$ ., then x equals

2005

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2004

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2003

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2001

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Solution

The correct option is
A
2005

$1.2003 + 2.2002 + 3.2001 + . . . . . + 2003.1 = 2003.334. x$

$1. (2004 - 1) + 2. (2004 - 2) + 3. (2004 - 3) + . . . . . + 2003. (2004 - 2003) = 2003.334. x$

$(1 + 2 + 3 + . . . . + 2003) 2004 - (1^{2} + 2^{2} + 3^{2} + . . . . . + 2003^{2}) = 2003.334. x$

$\frac{2003 \times 2004}{2} .2004 - 2003.4007.334$

$2003 \times 334 \times (3 \times 2004 - 4007) = 2003.334. x$

$2003 \times 334 \times 2005 = 2003.334. x$

$∴ x = 2005$

Suggest Corrections

Similar questions

1^{2} + 2^{2} + 3^{2} + . . . . . . + 2003^{2} = (2003) (4007) (334)

(1) (2003) + (2) (2002) + (3) (2001) + . . . . . + (2003) (1) = (2003) (334) (x) . .,

(1) (2003) + (2) (2002) + (3) (2001) + \dots + (2003) (1) = (2003) (334) (x),

[\frac{x}{5}]

S = 1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . - 2002^{2} + 2003^{2}

1^{2} - 2^{2} + 3^{3} - 4^{2} + . . . . - 2002^{2} + 2003^{2}

x = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . .,

y = \frac{1}{1^{2}} + \frac{3}{2^{2}} + \frac{1}{3^{2}} + \frac{3}{4^{2}} + . .

z = \frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . .,

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