8.2.2.2_Solution

8.2.2.2_Solution - 1 / 2 ⋅− 1 n 1 . The sum is obtained...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
INFS501 Homework 8.2.2.d Method 1 : Match the problem to the formula in Chapter 4 by reversing the summation order, then factoring out − 1 n : − 1 n  − 1 n 1 2 ... 2 n 3 2 n 2 2 n 1 2 n = − 1 n ⋅[ 1 2 4 ...  − 2 n ] , a geometric sequence with r = -2, = − 1 n k = 0 k = n − 2 k = − 1 n − 2 n 1 1 2 1 = 2 n 1 − 1 n 3 . Method 2 : Use the formula from class: In S = 2 n 2 n 1 2 n 2 ...  − 1 n 1 2  − 1 n , the first term is 2 n , r =− 1 / 2, and the next term after the final term would be
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 / 2 ⋅− 1 n 1 . The sum is obtained by subtracting the first term 2 n from what the missing term after the final term would be (i.e., from 1 / 2 ⋅− 1 n 1 , finally dividing by (common ratio - 1), i.e. dividing by − 3 / 2 = − 1 / 2 − 1: S = 1 / 2 ⋅− 1 n 1 − 2 n − 3 / 2 = 2 n 1 − 1 n 3 ....
View Full Document

This note was uploaded on 11/01/2010 for the course INFS 501 taught by Professor Ellis,w during the Spring '08 term at George Mason.

Ask a homework question - tutors are online