A 2012 Problem

Algebra Level 3

1005 2012 \frac{1005}{2012} 865 2012 \frac{865}{2012} 545 2012 \frac{545}{2012} 335 2012 \frac{335}{2012}

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1 solution

S = 1 2 × 5 + 1 5 × 8 + 1 8 × 11 + + 1 2009 × 2012 S = \dfrac{1}{2\times 5} + \dfrac{1}{5\times 8} + \dfrac{1}{8\times 11} + \ldots + \dfrac{1}{2009\times 2012}

S = n = 0 669 1 ( 3 n + 2 ) × ( 3 n + 5 ) = 1 3 n = 0 669 1 3 n + 2 1 3 n + 5 S = \displaystyle \sum_{n=0}^{669} \dfrac{1}{(3n+2)\times (3n+5)} = \dfrac{1}{3} \sum_{n=0}^{669} \dfrac{1}{3n+2} - \dfrac{1}{3n+5}

S = 1 3 ( 1 2 1 2012 ) S = \dfrac{1}{3}\left(\dfrac{1}{2} - \dfrac{1}{2012}\right)

S = 335 2012 S = \boxed{\dfrac{335}{2012}}

How did you solved from 1st step to the 2nd one....?½-1/2012?

Ronik Gandhi - 5 years, 9 months ago

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It is a telescopic sum.

Vishwak Srinivasan - 5 years, 9 months ago

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Can u explain pls....or share a link....

Ronik Gandhi - 5 years, 9 months ago

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@Ronik Gandhi When you expand the summation, you will realize that terms will cancel out in a pattern. Just try opening the summation.

Vishwak Srinivasan - 5 years, 9 months ago

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