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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

Option: 1

2005


Option: 2

2004


Option: 3

2003


Option: 4

2001


Answers (1)

best_answer

As we have learnt in 

1+2+3+4+------+n= \frac{n(n+1)}{2}

1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}

 

Now,

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

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

\sum_{r=1}^{2003} r (2003 -r+1 ) = (2003)(334) (x) \\\\ 2004 \cdot \sum_{r=1}^{2003} r - \sum_{r=1}^{2003} r^2 = (2003) (334) (x)\\\\ 2004 \cdot \left ( \frac{2003\cdot 2004}{2} \right ) - 2003 \cdot (4007\cdot 334) = (2003)(334)(x) \\\\ x = 2005

Posted by

HARSH KANKARIA

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