Suma z siódemką
: 12 sty 2025, o 19:58
Obliczyć \(\displaystyle{ \sum_{k=1}^{+ \infty} \frac{k^2}{7^{k-1}} }\).
\(\displaystyle{ y= \sum_{n=1}^{ \infty } x^n= \frac{1}{1-x} /'}\)
\(\displaystyle{ y'= \sum_{n=1}^{ \infty } nx^{n-1}=\left( \frac{1}{1-x} \right)'= \frac{1}{\left( 1-x\right)^2 }/ \cdot x }\)
\(\displaystyle{ y'x= \sum_{n=1}^{ \infty }nx^n= \frac{x}{\left( 1-x\right)^2 } /' }\)
\(\displaystyle{ y''x+y'= \sum_{n=1}^{ \infty }n^2x^{n-1}= \frac{1+x}{\left( 1-x\right)^3 } }\)
czyli niech:
\(\displaystyle{ h(x)= \sum_{n=1}^{ \infty }n^2x^{n-1}= \frac{1+x}{\left( 1-x\right)^3 }}\)
\(\displaystyle{ h\left( \frac{1}{7} \right) = \sum_{n=1}^{ \infty } \frac{n^2}{7^{n-1}} = \frac{1+ \frac{1}{7} }{\left( 1- \frac{1}{7} \right)^3 }= \frac{49}{27} }\)