{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

# Back then you were working in a base ten system here

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rmine which value of n produces the term 117. Setting an = 117, we get 2n − 1 = 117 or n = 59. Our ﬁnal answer is 59 1 + 3 + 5 + . . . + 117 = (2n − 1) n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives ‘−’ with the numerators to get 1 −1 1 −1 1 + ++ + ... + 1 2 3 4 117 The numerators, 1, −1, etc. can be described by the geometric sequence1 cn = (−1)n−1 for n ≥ 1, while the denominators are given by the arithmetic sequence2 dn = n for n−1 n ≥ 1. Hence, we get the formula an = (−1) for our terms, and we ﬁnd the lower and n upper limits of summation to be n = 1 and n = 117, respectively. Thus 111 1 1 − + − + −... + 234 117 117 = n=1 (−1)n−1 n 9 (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 10n for n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n − 1 zeros to the right of the decimal point before the 9, we...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online