2022/014) How do you prove that $\frac{2}{1!} + \frac{2+4}{2!} + \frac{2+4+6}{3!}+ \cdots = 3e$

 We have $n^{th}$ term 

$= \frac{2 + 4 + \cdots 2n }{n!} = \frac{n(n+1)}{n!}$

So then given exprsssion

$\frac{2}{1!} + \frac{2+4}{2!}  + \frac{2+4+6}{3!}+ \cdots$

$=\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$ putting the valut from above 

$=\sum_{n=1}^{\infty} \frac{n(n-1) + 2n }{n!}$ writing n(n+1) s n(n-1) + 2n

$=\sum_{n=1}^{\infty} (\frac{n(n-1)}{n!} + \frac{2n }{n!})$

$=\sum_{n=1}^{\infty} \frac{n(n-1)}{n!}  + \sum_{n=1}^{\infty} \frac{2n }{n!}$

$=\sum_{n=2}^{\infty} \frac{n(n-1)}{n!}  + \sum_{n=1}^{\infty} \frac{2n }{n!}$ we can convert the lowerr limit from n = 1 to n =2 because for n- 1 we have n(n-1) = 0 which can be dropped

$=\sum_{n=2}^{\infty} \frac{1}{(n-2)!}  + 2 \sum_{n=1}^{\infty} \frac{1 }{(n-1)!}$

$=\sum_{n=0}^{\infty} \frac{1}{n!}  + 2\sum_{n=0}^{\infty} \frac{1 }{n!}$ by changing the limit

 $=3 \sum_{n=0}^{\infty} \frac{1}{n!}$

= 3e by using expansion form for e