Chapter 12: Q. 12.19 (page 1170)

Expand the partial sum and find its value: $\sum_{k = 0}^{3} \frac{2}{k!}$

Short Answer

Expert verified

The value of the given partial sum is $\frac{16}{3} .$

Step by step solution

Step 1. Given information

$\begin{array}{l} Thepartialsumisgivenas \\ \sum_{k = 0}^{3} \frac{2}{k!} \\ Wehavetofindthevalueofthepartialsum. \end{array}$

Step 2. Expand the partial sum

$\begin{array}{l} Now, \\ Wesubstitutethevalues0, 1, 2, 3 inorder. \\ \Rightarrow \sum_{k = 0}^{3} \frac{2}{k!} = \frac{2}{0!} + \frac{2}{1!} + \frac{2}{2!} + \frac{2}{3!} \\ = \frac{2}{1} + \frac{2}{1} + \frac{2}{2 \times 1} + \frac{2}{3 \times 2 \times 1} [∵ n! = n (n - 1) (n - 2) ....] \\ = 2 + 2 + 1 + \frac{1}{3} \\ = 5 + \frac{1}{3} \\ = \frac{(5 \cdot 3) + 1}{3} \\ = \frac{16}{3} \end{array}$

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Expand the partial sum and find its value: $\sum_{k = 0}^{3} \frac{2}{k!}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Expand the partial sum

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Expand the partial sum and find its value:∑k=032k!

Short Answer

Step by step solution

Step 1. Given information

Step 2. Expand the partial sum

One App. One Place for Learning.

Most popular questions from this chapter

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Expand the partial sum and find its value: $\sum_{k = 0}^{3} \frac{2}{k!}$