Question

suma de la k=1 la n cu 1/k(k+1)(k+2) este egal cu ?

Answer 1

Răspuns:

Explicație pas cu pas:

$\dfrac{1}{k(k+1)(k+2)}=\dfrac{A}{k}+\dfrac{B}{k+1}+\dfrac{C}{k+2}\\1=A(k+1)(k+2)+Bk(k+2)+Ck(k+1)\\1=A(k^2+k+2k+2)+B\cdot k^2+B\cdot 2k+C\cdot k^2+C\cdot k\\1=A\cdot k^2+A\cdot 3k+2A+k^2(B+C)+k(2B+C)\\1=k^2(A+B+C)+k(3A+2B+C)+2A\\\begin{cases}A+B+C=0\\3A+2B+C=0\\2A=1\Rightarrow A=\dfrac{1}{2}\end{cases}$

$\begin{cases}\dfrac{1}{2}+B+C=0\\\dfrac{3}{2}+2B+C=0\end{cases}\Leftrightarrow \begin{cases}B+C=-\dfrac{1}{2}\\2B+C=-\dfrac{3}{2}\end{cases}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~------\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-B=1\Rightarrow B=-1\\-1+C=-\dfrac{1}{2}\Rightarrow C=\dfrac{1}{2}\\\texttt{Prin urmare:}\displaystyle\sum_{k=1}^n\dfrac{1}{k(k+1)(k+2)}=\sum_{k=1}^n\left(\dfrac{1}{2k}-\dfrac{1}{k+1}+\dfrac{1}{2(k+2)}\right)=$

$\displaystyle\dfrac{1}{2}\sum_{k=1}^n\left(\dfrac{1}{k}-\dfrac{2}{k+1}+\dfrac{1}{k+2}\right)=\\\dfrac{1}{2}\sum_{k=1}^n\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)+\dfrac{1}{2}\sum_{k=1}^n\left(\dfrac{1}{k+2}-\dfrac{1}{k+1}\right)=\\\dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)+\dfrac{1}{2}\left(\dfrac{1}{n+2}-\dfrac{1}{2}\right)=\\\dfrac{n}{2(n+1)}+\dfrac{1}{2}\cdot \dfrac{2-(n+2)}{2(n+2)}=\boxed{\dfrac{n}{2(n+1)}-\dfrac{n}{4(n+2)}}$