Tedious or elegant ?

Level 2

Let C r ( α ) = α ( α − 1 ) ( α − 2 ) . . . ( α − r + 1 ) r ! C_r(\alpha) = \displaystyle \frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-r+1)}{r!} . Then evaluate the definite integral

∫ 0 1 C 2015 ( − y − 1 ) ( 1 y + 1 + 1 y + 2 + . . . + 1 y + 2015 ) d y \displaystyle \int_{0}^{1} C_{2015}(-y-1)\left(\frac{1}{y+1}+\frac{1}{y+2}+...+\frac{1}{y+2015}\right) dy

*This is a modified problem....


The answer is -2015.

Pradeep Maurya by Pradeep Maurya , 32, India

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1 solution

Phineas Merrell
May 13, 2015

HOW THE HELL HAS NO ONE BUT ME GOTTEN THIS YET!!!!!!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Kudos to you! I guess many problem solvers havn't seen this problem yet...:)

Pradeep Maurya - 6 years, 1 month ago

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