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Jason99

Tes latex

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\[
\begin{align}
\sum_{n=1}^{k} \frac{n+2}{n! +  (n+1)! + (n+2)!} &= \sum_{n=1}^{k} \frac{n+2}{n! ( 1 + n + 1 + (n+1)(n+2)) } \\
 &= \sum_{n=1}^{k} \frac{n+2}{n! (n+2)(n+2) } \\
 &= \sum_{n=1}^{k} \frac{1}{\frac{(n+2)!}{n+1}} \\
 &= \sum_{n=1}^{k} \frac{n+1}{(n+2)!} \\ 
 &= \sum_{n=1}^{k} \frac{n+2 - 1 }{(n+2)!} \\ 
 &= \sum_{n=1}^{k} \frac{n+2}{(n+2)!} - \frac{1}{(n+2)!} \\
&=  \frac{1}{(n+1)!} - \frac{1}{(k+2)!}\\
\end{align}
\] 

 

\[ 

\begin{align}

\sum_{n=1}^{k} d_n &= \sum_{n=1}^{k} \frac{n(n+1)}{2} \\

&= \sum_{n=1}^{k} \left(\frac{n^2}{2} + \frac{n}{2} \right) \\

&= \frac{\sum_{n=1}^{k} n^2}{2} + \frac{\sum_{n=1}^{k} n}{2} \\

&= \frac{\frac{k(k+1)(2k+1)}{6}}{2} + \frac{\frac{k(k+1)}{2}}{2} \\

&=\frac{k(k+1)(2k+1)}{12} + \frac{k(k+1)}{4} \\

&=\frac{k(k+1)(k+2)}{6} \\

\end{align}

\]

Edited by Jason99

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