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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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???
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