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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 12: Q. 12.19 (page 1170)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information

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

02

Step 2. Expand the partial sum

$\begin{array}{l} Now, \\ Wesubstitutethevalues0, 1, 2, 3 inorder. \\ 鈬� {鈭�}_{k = 0}^{3} \frac{2}{k!} = \frac{2}{0!} + \frac{2}{1!} + \frac{2}{2!} + \frac{2}{3!} \\ = \frac{2}{1} + \frac{2}{1} + \frac{2}{2 脳 1} + \frac{2}{3 脳 2 脳 1} [鈭� n! = n (n 鈭� 1) (n 鈭� 2) ....] \\ = 2 + 2 + 1 + \frac{1}{3} \\ = 5 + \frac{1}{3} \\ = \frac{(5 鈰� 3) + 1}{3} \\ = \frac{16}{3} \end{array}$

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