Functions?

**fardeen_gen** · April 25th 2009, 09:58 PM

If f(x) = {k^x/(k^x + √k)} (k >0) then prove that:
2n -1
∑ 2f(r/2n) = (2n - 1)
r = 1

**Media_Man** · May 27th 2009, 01:09 PM

Define $f(x)=\frac{k^x}{k^x+k^{1/2}}$ for some $k>0$

Define $a_r=f(\frac r{2n})$

Show that $2\sum_{r=1}^{2n-1} a_r = 2n-1$ for all $n>1$ , $k>0$

Rearrange $a_1+a_2+a_3+...+a_{2n-2}+a_{2n-1}$ like so: $a_{n}+(a_{n-1}+a_{n+1})+(a_{n-2}+a_{n+2})+...+(a_1+a_{2n-1})$

So $2\sum_{r=1}^{2n-1} a_r= 2a_n+2\sum_{p=1}^{n-1}(a_{n-p}+a_{n+p})$

Now $2a_n=2f(\frac n{2n})=2f(\frac12)=2\frac{k^{1/2}}{k^{1/2}+k^{1/2}}=1$

And $a_{n-p}+a_{n+p}= \frac{k^{(n-p)/2n}}{k^{(n-p)/2n}+k^{1/2}}+\frac{k^{(n+p)/2n}}{k^{(n+p)/2n}+k^{1/2}}$

$a_{n-p}+a_{n+p}=\frac{(k^{(n-p)/2n})(k^{(n+p)/2n}+k^{1/2})+(k^{(n+p)/2n})(k^{(n-p)/2n}+k^{1/2})}{(k^{(n-p)/2n}+k^{1/2})(k^{(n+p)/2n}+k^{1/2})}=1$

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

Therefore, $2a_n+2\sum_{p=1}^{n-1}(a_{n-p}+a_{n+p})=1+2(n-1)=2n-1$

QED

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

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