If f(x) = {k^x/(k^x + √k)} (k >0) then prove that:

2n -1

∑ 2f(r/2n) = (2n - 1)

r = 1

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- Apr 25th 2009, 09:58 PMfardeen_genFunctions?
If f(x) = {k^x/(k^x + √k)} (k >0) then prove that:

2n -1

∑ 2f(r/2n) = (2n - 1)

r = 1 - May 27th 2009, 01:09 PMMedia_ManAchtung Baby!
Define for some

Define

Show that for all ,

Rearrange like so:

So

Now

And

(Multiplying out, the numerator and denominator are equal for all n,p,k.)

Therefore,

QED

**Where do you get these crazy problems???