Question

Suma de la k = 1 până la n : $\frac{k+2}{k(k+1)} 2^{-k}$

Answer 1

$\displaystyle \sum\limits^n_{k=1} \frac{k+2}{k(k+1)} \cdot 2^{-k}=\sum\limits^n_{k=1}\left( \frac{1}{k(k+1)}+ \frac{1}{k} \right) \cdot \frac{1}{2^k}= \\ \\ =\sum\limits^n_{k=1} \left( \frac{1}{k}- \frac{1}{k+1}+ \frac{1}{k} \right) \cdot \frac{1}{2^k}=\sum\limits^n_{k=1} \left(\frac{1}{k \cdot 2^{k-1}}-\frac{1}{(k+1) \cdot 2^k} \right)=$
[tex]\displaystyle =\frac{1}{1 \cdot 2^0}- \frac{1}{2 \cdot 2^1}+ \frac{1}{2 \cdot 2^1}- \frac{1}{3 \cdot 2^2}+ \frac{1}{3 \cdot 2^2}- \frac{1}{4 \cdot 2^3}+...+ \\ \\ +\frac{1}{n \cdot 2^{n-1}}- \frac{1}{(n+1)\cdot 2^n}= (se~reduc~termenii,~ramanand~ \\ \\ numai~primul~si~ultimul~-~telescopare) \\ \\ =\frac{1}{1 \cdot 2^0}- \frac{1}{(n+1) \cdot 2^n }= \\ \\ =1- \frac{1}{(n+1) \cdot 2^n}. \\ \\ Deci~\boxed{\sum\limits^n_{k=1} \frac{k+2}{k(k+1)} \cdot 2^{-k} = 1- \frac{1}{(n+1) \cdot 2^n}}~. [/tex]

Answer 2

Trebuie descompus termenul general, si se reduc in afara de doi termeni.