Calculus 8th Edition

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

Appendix E - Sigma Notation - E Exercises: 32

Answer

$$\sum_{i=1}^n(3+2i)^2=\frac{1}{3}n(4n^2+24n+47)$$

Work Step by Step

Expanding the expression under the sum and then summing term by term, with formulas from theorem 3 we get, step by step: $$A=\sum_{i=1}^n(3+2i)^2=\sum_{i=1}^n(9+12i+4i^2)=\\ =\sum_{i=1}^n9+12\sum_{i=1}^ni+4\sum_{i=1}^ni^2=\\= 9n+12\frac{n(n+1)}{2}+4\frac{n(n+1)(2n+1)}{6}=\\= 9n+6n(n+1)+\frac{2}{3}n(n+1)(2n+1)=\\= n\left(9+6(n+1)+\frac{2}{3}(n+1)(2n+1)\right)=\\= n\left(9+(n+1)\left(6+\frac{2}{3}(2n+1)\right)\right)=\\= n\left(9+(n+1)\left(\frac{4n+20}{3}\right)\right)=\\= n\left(9+\frac{4(n+1)(n+5)}{3}\right)=\\=\frac{1}{3}n(4n^2+24n+47).$$
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