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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. 12.20 (page 1170)

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

Short Answer

Expert verified

The value of the partial sum ${鈭�}_{k = 0}^{3} \frac{3}{k!}$ is 8.

Step by step solution

01

Step 1. Given information

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

02

Step 2. Expand the partial sum

$\begin{array}{l} Now, \\ Wesubstitutethevalues0, 1, 2, 3 inorder. \\ 鈬� {鈭�}_{k = 0}^{3} \frac{3}{k!} = \frac{3}{0!} + \frac{3}{1!} + \frac{3}{2!} + \frac{3}{3!} \\ = \frac{3}{1} + \frac{3}{1} + \frac{3}{2 脳 1} + \frac{3}{3 脳 2 脳 1} [鈭� n! = n (n 鈭� 1) (n 鈭� 2) ....] \\ = 3 + 3 + \frac{3}{2} + \frac{1}{2} \\ = 6 + \frac{4}{2} \\ = 6 + 2 \\ = 8 \end{array}$

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