Question

Expand the partial sum and find its value type your answer as an improper fraction:sigma (sum) starting value (k) at 0 ending value 3, function is 4/k!The sum is Answer

Answer 1

Given:

`\sum_{k\mathop{=}0}^3(4)/(k!)`

This implies that the value of k starts from 0 and ends at 3.

Thus, the partial sum is expanded to be

`(4)/(0!)+(4)/(1!)+(4)/(2!)+(4)/(3!)`

Where

`\begin{gathered} (4)/(0!)=(4)/(1)=4 \\ (4)/(1!)=(4)/(1)=4 \\ (4)/(2!)=(4)/(2)=2 \\ (4)/(3!)=(4)/(6)=(2)/(3) \end{gathered}`

Thus, we have

`\begin{gathered} 4+4+2+(2)/(3) \\ =(32)/(3) \end{gathered}`

Hence, the sum is

`(32)/(3)`