JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

{1^2} + {2^2} + {3^2} + \ldots + {2003^2} = \left( {2003} \right)\left( {4007} \right)\left( {334} \right)

and,

1 \times 2003 + 2 \times 2002 + 3 \times 2001 + \ldots + 2003 \times 1 = \left( {2003} \right)\left( {334} \right)x,

then

x =

View answer and explanation

Correct option: A

2005

Option A (2005) is correct because by manipulating the given equations and simplifying, we can solve for x and find its value to be 2005.

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