Question

evaluate

Answer 1

\[f(x) = \frac{ 2 }{2+ 4^{x} }\]
the value of
\[f(\frac{ 1 }{ 2014 }) + f(\frac{ 2 }{ 2014 }) + f(\frac{ 3 }{ 2014 }) + ... + f(\frac{ 2013 }{ 2014 })\]

Answer 2

please give me a hint, advices or something :)

Answer 3

i know i have to subtitute x = 1/2014, 2/2014, ... till 2013/2014
but it looks horor to me

Answer 4

maybe lets prove this first :
\[f(1/2014) + f(2013/2014) = f(2/2014) + f(2012/2014) = \cdots = 1\]

Answer 5

prove :

\[f(k/2014) + f((2014-k)/2014) = 1\]

Answer 6

with k = 1 till 2013 ?

Answer 7

we just the first half i think as we're pairing up

Answer 8

\[\large f(k/2014) + f((2014-k)/2014) = \dfrac{2}{2+4^{\frac{k}{2014}}} + \dfrac{2}{2+4^{\frac{2014-k}{2014}}}\]

Answer 9

\[\large = 2\left(\dfrac{2+4^{\frac{2014-k}{2014}} + 2+4^{\frac{k}{2014}}}{\left(2+4^{\frac{2014-k}{2014}} \right)\left( 2+4^{\frac{k}{2014}}\right)} \right)\]

\[\large = 2\left(\dfrac{2+4^{\frac{2014-k}{2014}} + 2+4^{\frac{k}{2014}}}{2\left( 2+4^{\frac{2014-k}{2014}} + 2+4^{\frac{k}{2014}}\right)} \right)\]


\[\large = 1\]

Answer 10

You can evaluate the middle value when \(k = 1007\) directly :

\[f(1007/2014) = \dfrac{2}{2+4^{\frac{1}{2}}} = \dfrac{1}{2}\]

Answer 11

overall the sum would be :

\(\large \dfrac{2012}{2} + \dfrac{1}{2}\)

Answer 12

I was thinking the same^^ but I was doubting whether my thought about "pairing up of terms" seems true,couldn't prove that part but it should be done like this only :D

Answer 13

one constant term i.e the middle term and twice of sum of first half of the terms

Answer 14

oh, yes. there is a middle term there. i just confused how to simplify the multiplication of the denominator part. but i got it now, thanks a ton @ganeshie8 :)

Answer 15

np :)