1. Tardigrade
  2. Question
  3. Mathematics
  4. Let Sn=1 ⋅(n-1)+2 ⋅(n-2)+3 ⋅(n-3)+ ldots+ ( n -1) ⋅ 1, n ≥ 4 The sum displaystyle∑n=4∞((2 Sn/n !)-(1/(n-2) !)) is equal to :

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Q. Let $S_{n}=1 \cdot(n-1)+2 \cdot(n-2)+3 \cdot(n-3)+\ldots+$ $( n -1) \cdot 1, n \geq 4$
The sum $\displaystyle\sum_{n=4}^{\infty}\left(\frac{2 S_{n}}{n !}-\frac{1}{(n-2) !}\right)$ is equal to :

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Solution:

Let $T _{ r }= r ( n - r )$
$T _{ r }= nr - r ^{2}$
$\Rightarrow S _{ n }=\displaystyle\sum_{ r =1}^{ n } T _{ r }=\displaystyle\sum_{ r =1}^{ n }\left( nr - r ^{2}\right)$
$S _{ n }=\frac{ n \cdot( n )( n +1)}{2}-\frac{ n ( n +1)(2 n +1)}{6}$
$S _{ n }=\frac{ n ( n -1)( n +1)}{6}$
Now $\displaystyle\sum_{ r =4}^{\infty}\left(\frac{2 S _{ n }}{ n !}-\frac{1}{( n -2) !}\right)$
$=\displaystyle\sum_{ r =4}^{\infty}\left(2 \cdot \frac{ n ( n -1)( n +1)}{6 \cdot n ( n -1)( n -2) !}-\frac{1}{( n -2) !}\right)$
$=\displaystyle\sum_{ r =4}^{\infty}\left(\frac{1}{3}\left(\frac{ n -2+3}{( n -2) !}\right)-\frac{1}{( n -2) !}\right)$
$=\frac{1}{( n -3) !}=\frac{1}{3}( e -1)$