Question

Hello,Can you help me with question number 9 in the picture? It says find the indicated sum

Answer 1

Answer:
`\sum_{i\mathop{=}0}^4(-1)^(i+1)i!\text{ = -20}`Step-by-step explanation:

The given summation is:

`\sum_{i\mathop{=}0}^4(-1)^(i+1)i!`

For i = 0

`\begin{gathered} (-1)^(0+1)(0!) \\ =(-1)^1*1 \\ =-1 \end{gathered}`

For i = 1

`\begin{gathered} (-1)^(1+1)(1!) \\ \\ =(-1)^2*1 \\ \\ =1 \end{gathered}`

For i = 2

`\begin{gathered} (-1)^(2+1)2! \\ \\ (-1)^3*2! \\ \\ =-1*2*1 \\ \\ =-2 \end{gathered}`

For i = 3

`\begin{gathered} (-1)^(3+1)(3!) \\ \\ (-1)^4(3!) \\ \\ =1*3*2*1 \\ \\ =6 \end{gathered}`

For i = 4

`\begin{gathered} (-1)^(4+1)(4!) \\ \\ =(-1)^5*4! \\ \\ =-1*4*3*2*1 \\ \\ =-24 \end{gathered}`

Therefore, the required sum is:

`\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}`