Answer:[tex]\sum_{i\mathop{=}0}^4(-1)^{i+1}i!\text{ = -20}[/tex]Explanation:
The given summation is:
[tex]\sum_{i\mathop{=}0}^4(-1)^{i+1}i![/tex]
For i = 0
[tex]\begin{gathered} (-1)^{0+1}(0!) \\ =(-1)^1\times1 \\ =-1 \end{gathered}[/tex]
For i = 1
[tex]\begin{gathered} (-1)^{1+1}(1!) \\ \\ =(-1)^2\times1 \\ \\ =1 \end{gathered}[/tex]
For i = 2
[tex]\begin{gathered} (-1)^{2+1}2! \\ \\ (-1)^3\times2! \\ \\ =-1\times2\times1 \\ \\ =-2 \end{gathered}[/tex]
For i = 3
[tex]\begin{gathered} (-1)^{3+1}(3!) \\ \\ (-1)^4(3!) \\ \\ =1\times3\times2\times1 \\ \\ =6 \end{gathered}[/tex]
For i = 4
[tex]\begin{gathered} (-1)^{4+1}(4!) \\ \\ =(-1)^5\times4! \\ \\ =-1\times4\times3\times2\times1 \\ \\ =-24 \end{gathered}[/tex]
Therefore, the required sum is:
[tex]\begin{gathered} \sum_{i\mathop{=}0}^4(-1)^{i+1}(i!)=-1+(1)+(-2)+6+(-24) \\ \\ \sum_{i\mathop{=}0}^4(-1)^{i+1}(i!)=-20 \end{gathered}[/tex]
