Suma z potęgami dziewiątki
: 11 lip 2015, o 14:49
Ile to jest \(\displaystyle{ \sum_{0 \leq k: \leq 100 \ NWD \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)}\) gdy \(\displaystyle{ f \left( x \right) = \frac{9^x}{3+9^x}}\) ?
\(\displaystyle{ f\left( \frac{k}{100} \right) = \frac{9^{ \frac{k}{100} }}{3+9^{ \frac{k}{100} }}=\frac{3^{ \frac{k}{50} }}{3+3^{ \frac{k}{50} }} }\)
\(\displaystyle{ \sum_{(k,100)=1}^{} f\left( \frac{k}{100} \right) = \sum_{k<50}^{} \frac{3^{ \frac{k}{50} }}{3+3^{ \frac{k}{50} }}+\sum_{k>50}^{} \frac{3^{ \frac{k}{50} }}{3+3^{ \frac{k}{50} }}=}\)
\(\displaystyle{ =\sum_{k<50}^{} \frac{3 \cdot 3^{ \frac{k-50}{50} }}{3+3 \cdot 3^{ \frac{k-50}{50} }}+\sum_{k<50}^{}\frac{ 3^{ 1+\frac{k}{50} }}{3+ 3^{1+ \frac{k}{50} }}=}\)
\(\displaystyle{ =\sum_{k<50}^{} \frac{ 3^{ \frac{k-50}{50} }/ \cdot 3^{ \frac{50-k}{50} }}{1+ 3^{ \frac{k-50}{50} }/ \cdot 3^{ \frac{50-k}{50} }}+\sum_{k<50}^{}\frac{ 3^{ \frac{k}{50} }}{1+ 3^{\frac{k}{50} }}=}\)
\(\displaystyle{ = \sum_{k<50}^{} \frac{1}{1+3^{ \frac{50-k}{50} }}+\sum_{k<50}^{}\frac{ 3^{ \frac{k}{50} }}{1+ 3^{\frac{k}{50} }} =\sum_{k<50}^{}\frac{ 1}{1+ 3^{\frac{k}{50} }}+\sum_{k<50}^{}\frac{ 3^{ \frac{k}{50} }}{1+ 3^{\frac{k}{50} }}=\sum_{k<50}^{}\frac{ 1+ 3^{ \frac{k}{50} }}{1+ 3^{\frac{k}{50} }}= \sum_{(k,50)=1}^{} 1=\\=\varphi(50)=\varphi(2 \cdot 5^2)=\varphi(2) \cdot \varphi(5^2)=1 \cdot 5 \cdot 4=20}\)