Past, Present, or Future?

Algebra Level 3

$\sum_{k=0}^{2014}\left(\frac{2}{2+e^{2k\pi i/2015}}\right)$

Let $S$ denote the value of the summation above. Which of the answer choices must be true?

Bonus question : Can you generalize this? If $\displaystyle S_n=\sum_{k=0}^{n-1}\left(\frac{2}{2+e^{2k\pi i/n}}\right)$ , what is the relationship between $S_n$ and $n$ ?

S

is undefined

S=2015

S>2015

S<2015

1 solution

Satvik Pandey
Oct 20, 2015

Here ${ e }^{ \frac { 2k\pi i }{ n } }\quad$ where 'n' can take integral values form $0-2014$ are simply roots of ${ z }^{ 2015 }=1$

So ${ z }^{ 2015 }-1=\prod _{ i=1 }^{ 2015 }{ (x-{ \omega }_{ i }) }$

$\log { { (z }^{ 2015 }-1) } =\log { \left( \prod _{ i=1 }^{ 2015 }{ (x-{ \omega }_{ i }) } \right) }$

$\frac { 2015{ z }^{ 2014 } }{ { z }^{ 2015 }-1 } =\sum _{ i=1 }^{ 2015 }{ \frac { 1 }{ (x-{ \omega }_{ i }) } }$

$\frac { 2015{ (-2) }^{ 2014 } }{ (-2)^{ 2015 }-1 } \times 2=-\sum _{ i=1 }^{ 2015 }{ \frac { 2 }{ (2+{ \omega }_{ i }) } }$

$\sum _{ i=1 }^{ 2015 }{ \frac { 2 }{ (2+{ \omega }_{ i }) } } =2015\left( \frac { { 2 }^{ 2015 } }{ { 2 }^{ 2015 }+1 } \right) <2015$ .

Yes, exactly... very clear solution (upvote)! What happens when $n$ is even?

The number is "in the past", $<2015$ ... but just barely.

Otto Bretscher - 5 years, 7 months ago

Thank you, sir! The number is " in future" if 'n' is even. :D

The value of ${ S }_{ n }$ is $\frac { n{ 2 }^{ n } }{ { 2 }^{ n }+1 }$ when 'n' is odd and $\frac { n{ 2 }^{ n } }{ { 2 }^{ n }-1 }$ when it is even.

satvik pandey - 5 years, 7 months ago

Past, Present, or Future?

1 solution

0 pending reports