find-the-following-sum-k-1-n-C-n-k-x-k-k-1-

Question Number 101078 by Coronavirus last updated on 30/Jun/20

f i n d t h e f o l l o w i n g s u m

\underset{k = 1}{\sum^{n}} C_{n}^{k} x^{- k} (k - 1)!

Answered by abdomathmax last updated on 01/Jul/20

f (u) = \sum_{k = 1}^{n} C_{n}^{k} (k - 1)! u^{k} (herr u = x^{- 1}) \Rightarrow

f (u) = \sum_{k = 1}^{n} \frac{n!}{k! (n - k)!} (k - 1)! u^{k}

= n! \sum_{k = 1}^{n} \frac{u^{k}}{k (n - k)!} \Rightarrow f^{^{'}} (u) = n! \sum_{k = 1}^{n} \frac{u^{k - 1}}{(n - k)!}

=_{n - k = p} n! \sum_{0}^{n - 1} \frac{u^{n - p - 1}}{p!}

= n! u^{n - 1} \sum_{p = 0}^{n - 1} \frac{{(u^{- 1})}^{p}}{p!} we have

\sum_{p = 0}^{\infty} \frac{{(u^{- 1})}^{p}}{p!} = e^{u^{- 1}} = \sum_{p = 0}^{n - 1} \frac{{(u^{- 1})}^{p}}{p!} + \sum_{p = n}^{\infty} \frac{{(u^{- 1})}^{p}}{p!}

\Rightarrow \sum_{p = 0}^{n - 1} \frac{{(u^{- 1})}^{p}}{p!} = e^{\frac{1}{u}} - \sum_{p = n}^{\infty} \frac{u^{- p}}{p!}

\Rightarrow f^{^{'}} (u) = n! u^{n - 1} (e^{\frac{1}{u}} - \sum_{p = n}^{\infty} \frac{u^{- p}}{p!}) \Rightarrow

f (u) = n! \int u^{n - 1} e^{\frac{1}{u}} du - n! \sum_{p = n}^{\infty} \frac{1}{p!} \int u^{n - 1 - p} du + c

be continued \dots .