If f(x) = fracx^22x^2 + sqrt(2), x in mathbbR, then sum _k=1^81 f ( (k / 82) ) is equal to

f( x)=frac2^ x2^ x+sqrt(2) f( x)+ f(1- x)=frac2^ x2^ x+sqrt(2)+frac2^1- x2^1- x+sqrt(2) =frac2^ x2^ x+sqrt(2)+frac22+sqrt(2) 2^ x =frac2^ x+sqrt(2)2^ x+sqrt(2)=1 Now, \ sum _ k=1^81 f(frac k82)= f((1 / 82))+ f((2 / 82))+ ... ... + f((81 / 82)) = f((1 / 82))+ f((2 / 82))+ ... ... + f(1-(2 / 82))+ f(1-(1 / 82)) =[ f((1 / 82))+ f(1-(1 / 82))]+[ f((2 / 82))+ f(1-(2 / 82))]+ ... .40 cases + f((41 / 82)) =(1+1+ ... 40 times )+frac2^1 / 22^1 / 2+2^1 / 2 40+(1 / 2)=(81 / 2)

If fx frac x22x2 2 x R then sumk181 f frac k82 is equal... | (Mathematics)

$\mathrm{f}(\mathrm{x})=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}+\sqrt{2}} \\\\$

$\mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}+\sqrt{2}}+\frac{2^{1-\mathrm{x}}}{2^{1-\mathrm{x}}+\sqrt{2}} \\\\$

$=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^{\mathrm{x}}}\\$ = $\frac{2^{\mathrm{x}}+\sqrt{2}}{2^{\mathrm{x}}+\sqrt{2}}=1 \\\\$

Now, $\\ \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\$

$=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right) \\$

$=\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots .40$ { cases }+ $\mathrm{f}\left(\frac{41}{82}\right) \\$

$=(1+1+\ldots 40 \text { times })+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\$

$40+\frac{1}{2}$ = $\frac{81}{2}$

Explanation

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