obliczyć sume szeregów

[olciaa]

\(a) \sum_{ n=1 }^{ \infty } \frac{2^n+4^n+6^n}{8^n}\)
\(b) \sum_{n=1}^{ \infty } \frac{(-1)^{n+1}}{3^n}\)
\(c) \sum_{n=1}^{ \infty } \frac{1}{n(n+3)}\)

[radagast]

\(a) \sum_{ n=1 }^{ \infty } \frac{2^n+4^n+6^n}{8^n}\)

\(\sum_{ n=1 }^{ \infty } \frac{2^n+4^n+6^n}{8^n} =\sum_{ n=1 }^{ \infty } \frac{2^n}{8^n} +\frac{4^n}{8^n} +\frac{6^n}{8^n} = \sum_{ n=1 }^{ \infty } \left( \frac{1}{4} \right)^n +\left( \frac{1}{2} \right)^n+\left( \frac{3}{4} \right)^n =\sum_{ n=1 }^{ \infty } \left( \frac{1}{4} \right)^n +\sum_{ n=1 }^{ \infty }\left( \frac{1}{2} \right)^n+\sum_{ n=1 }^{ \infty }\left( \frac{3}{4} \right)^n =
\frac{ \frac{1}{4} }{1- \frac{1}{4} } + \frac{ \frac{1}{2} }{1- \frac{1}{2} }+ \frac{ \frac{3}{4} }{1- \frac{3}{4} }= \frac{1}{3} +1+3= \frac{13}{3}\)

[octahedron]

\(b)
\sum_{n=1}^{ \infty } \frac{(-1)^{n+1}}{3^n}=\frac{1}{3}-\frac{1}{9}+\frac{1}{27}-\frac{1}{81}+...=\(\frac{1}{3}+\frac{1}{27}+...\)-\(\frac{1}{9}+\frac{1}{81}+...\)=\frac{\frac{1}{3}}{1-\frac{1}{9}}-\frac{\frac{1}{9}}{1-\frac{1}{9}}=\frac{1}{4}\)

[octahedron]

\(c)
\sum_{n=1}^{ \infty } \frac{1}{n(n+3)}=\sum_{n=1}^{ \infty }\frac{1}{3}\( \frac{1}{n}-\frac{1}{n+3}\)=\frac{1}{3}\(1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\frac{1}{5}-\frac{1}{8}+\frac{1}{6}-\frac{1}{9}+...\)=
=\frac{1}{3}\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...\)-\frac{1}{3}\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...\)=\frac{1}{3}\(1+\frac{1}{2}+\frac{1}{3}\)=\frac{11}{18}\)