It took me one and a half hours to show this formula X(
We have replaced k - l by k in the first line and have changed how to take sums by the formula k + j = m in the fourth line. Finally, this formula enable us to compute
A formula in the combination theory
I was asked to show the following formula in the last part time job:
!} \sum _{j = 0}^{n - k} \frac{(-1)^j}{j!} = 1, \textrm{ for } l = 0, \ldots, n. "\sum_{k = l}^n \frac{1}{(k - l)!} \sum _{j = 0}^{n - k} \frac{(-1)^j}{j!} = 1, \textrm{ for } l = 0, \ldots, n.")
It took me one and a half hours to show this formula X(
}} \\ = & \displaystyle{\sum_{0 \leq k \leq n - l} \frac{1}{k!} \sum_{0 \leq j \leq n - k - l} \frac{(-1)^j}{j!} } \\ = & \displaystyle{\sum_{0 \leq k \leq N} \sum_{0 \leq j \leq N - k} \frac{(-1)^j}{k! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \sum_{0 \leq j \leq N - m} \frac{(-1)^j}{(m - j)! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \frac{1}{m!} \sum_{0 \leq j \leq N - m} \frac{m!}{(m - j)! j!}} (-1)^j \\ = & \displaystyle{1 + \sum_{1 \leq m \leq N} \frac{1}{m!} \{ 1 + (-1) \}^m } \\ = & \displaystyle{1} \\ \end{array} "\begin{array}{cl} & \displaystyle{\textrm{(L.H.S.)}} \\ = & \displaystyle{\sum_{0 \leq k \leq n - l} \frac{1}{k!} \sum_{0 \leq j \leq n - k - l} \frac{(-1)^j}{j!} } \\ = & \displaystyle{\sum_{0 \leq k \leq N} \sum_{0 \leq j \leq N - k} \frac{(-1)^j}{k! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \sum_{0 \leq j \leq N - m} \frac{(-1)^j}{(m - j)! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \frac{1}{m!} \sum_{0 \leq j \leq N - m} \frac{m!}{(m - j)! j!}} (-1)^j \\ = & \displaystyle{1 + \sum_{1 \leq m \leq N} \frac{1}{m!} \{ 1 + (-1) \}^m } \\ = & \displaystyle{1} \\ \end{array}")
We have replaced k - l by k in the first line and have changed how to take sums by the formula k + j = m in the fourth line. Finally, this formula enable us to compute
^j}{j!} = \begin{cases} 2 & \textrm{ if } n \geq 2 \\ 1 & \textrm{ if } n = 1. \\ \end{cases} "\sum_{k = 0}^n \frac{k^2}{k!} \sum _{j = 0}^{n - k} \frac{(-1)^j}{j!} = \begin{cases} 2 & \textrm{ if } n \geq 2 \\ 1 & \textrm{ if } n = 1. \\ \end{cases}")
It took me one and a half hours to show this formula X(
We have replaced k - l by k in the first line and have changed how to take sums by the formula k + j = m in the fourth line. Finally, this formula enable us to compute
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