Question ID: #431
If $f(x) = \frac{2^x}{2^x + \sqrt{2}}$, $x \in R$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to:
- (1) 41
- (2) $\frac{81}{2}$
- (3) 82
- (4) $81\sqrt{2}$
Solution:
Given $f(x) = \frac{2^x}{2^x + \sqrt{2}}$.
Consider the sum $f(x) + f(1-x)$:
$$ f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2}{2 + \sqrt{2} \cdot 2^x} = \frac{\sqrt{2}}{\sqrt{2} + 2^x} $$
$$ f(x) + f(1-x) = \frac{2^x + \sqrt{2}}{2^x + \sqrt{2}} = 1 $$
Let $S = \sum_{k=1}^{81} f\left(\frac{k}{82}\right)$.
The terms can be paired as $f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = f\left(\frac{k}{82}\right) + f\left(1 – \frac{k}{82}\right) = 1$.
Total terms = 81.
Number of pairs = 40 (summing to 40).
Middle term (at $k=41$) = $f\left(\frac{41}{82}\right) = f\left(\frac{1}{2}\right)$.
$$ f(1/2) = \frac{2^{1/2}}{2^{1/2} + \sqrt{2}} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} $$
$$ S = 40 + \frac{1}{2} = \frac{81}{2} $$
Ans. (2)
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