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Với $x = \sin^2 \frac{k \pi}{100}$ thì $1-2x =\cos\frac{k \pi}{50}\Rightarrow f( \sin^2 \frac{k \pi}{100})=\frac{1}{1+2^{\cos\frac{k \pi}{50}}}$
Ta tính
$f( \sin^2 \frac{k \pi}{100})+f( \sin^2 \frac{(50-k) \pi}{100})=\frac{1}{1+2^{\cos\frac{k \pi}{50}}}+\frac{1}{1+2^{\cos\frac{(50-k) \pi}{50}}}$
$=\frac{1}{1+2^{\cos\frac{k \pi}{50}}}+\frac{1}{1+2^{\cos(\pi -\frac{k \pi}{50})}}=\frac{2+2^{\cos\frac{k \pi}{50}}+2^{\cos(\pi -\frac{k \pi}{50})}}{2+2^{\cos\frac{k \pi}{50}}+2^{\cos(\pi -\frac{k \pi}{50})}}=1$
Do đó $S=\sum_{k=1}^{24}(f( \sin^2 \frac{k \pi}{100})+f( \sin^2 \frac{(50-k) \pi}{100}))+f( \sin^2 \frac{25 \pi}{100})+f( \sin^2 \frac{50 \pi}{100})$
$=24+\frac{1}{2}+\frac{2}{3}=\frac{151}{6}$
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