Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix E - Sigma Notation - E Exercises - Page A38: 41

Answer

a) $n^4$ b) $5^{100}-1$ c)$\frac{97}{300}$ d) $a_n-a_0$

Work Step by Step

a) Expanding the given sum gives us $\sum_{i=1}^n [i^4−(i−1)^4]=(1^4−0^4)+(2^4−1^4)+(3^4−2^4)+\cdots +[n^4−(n−1)^4]$. We simplify the result by removing the terms that cancel out. Thus we obtain $\sum_{i=1}^n [i^4−(i−1)^4]=-0^4+n^4=n^4$. b) As in (a), expanding the sum gives us $\sum_{i=1}^{100} (5^i−5^{i−1})=(5^1−5^0)+(5^2−5^1)+(5^3−5^2)+\cdots +(5^{100}−5^{99})$ We then simplify the result by removing the terms that cancel out. Hence, $\sum_{i=1}^{100} (5^i−5^{i−1})=−5^0+5^{100} = 5^{100}−1$. c) We expand the sum and obtain $\sum_{i=3}^{99} \left(\frac{1}{i}−\frac{1}{i+1}\right) = \left(\frac{1}{3}−\frac{1}{4}\right)+\left(\frac{1}{4}−\frac{1}{5}\right)+\left(\frac{1}{5}−\frac{1}{6}\right)+\left(\frac{1}{6}−\frac{1}{7}\right)+\cdots +\left(\frac{1}{99}−\frac{1}{100}\right)$ We simplify the result by taking out terms that cancel out. Thus, $\sum_{i=3}^{99} \left(\frac{1}{i}−\frac{1}{i+1}\right)=\frac{1}{3}−\frac{1}{100} = \frac{100−3}{300} = \frac{97}{300}$. d) As in the above solutions, we first expand the given sum. That is, $\sum_{i=1}^{n} (a_i−a_{i−1}) = (a_1−a_0)+(a_2−a_1)+(a_3−a_2)+(a_4−a_3)+\cdots +(a_n−a_{n−1})$. Then we remove those terms that cancel out. Hence we obtain $\sum_{i=1}^{n} (a_i−a_{i−1})=−a_0+a_n = a_n−a_0$.
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