Evaluate each finite geometric series. 7.n-12-64n=1

Given that ∑n=18-6412n-1To find : finite geometric series. Solution : ∑n=18-6412n-1= -64 ∑n=1812n-1…

Answered: Σ n-1 -64 n=1

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

Evaluate each finite geometric series.

Expert Solution

Step 1

$G i v e n t h a t \sum_{n = 1}^{8} - 64 {(\frac{1}{2})}^{n - 1} T o f i n d : f i n i t e g e o m e t r i c s e r i e s . S o l u t i o n : \sum_{n = 1}^{8} - 64 {(\frac{1}{2})}^{n - 1} = - 64 \sum_{n = 1}^{8} {(\frac{1}{2})}^{n - 1} = - 64 [0 + {(\frac{1}{2})}^{1} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + . . . . . + {(\frac{1}{2})}^{7}] = - 64 [{(\frac{1}{2})}^{1} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + . . . . . + {(\frac{1}{2})}^{7}] = - 64 [(\frac{1}{2}) \frac{{(\frac{1}{2})}^{7} - 1}{(\frac{1}{2}) - 1}] = - 64 [(\frac{1}{2}) \frac{{(2)}^{- 7} - 1}{- (\frac{1}{2})}] = - 64 [(\frac{1}{2}) \frac{{(2)}^{- 7} - 1}{- (\frac{1}{2})}] = 64 [{(2)}^{- 7} - 1] = 2^{6} [{(2)}^{- 7} - 1] = 2^{6 - 7} - 2^{6} = 2^{- 1} - 2^{6} = \frac{1}{2} - 2^{6}$