I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 10.3 : Series - Basics
1. Perform an index shift so that the following series starts at \(n = 3\).
\[\sum\limits_{n = 1}^\infty {\left( {n{2^n} - {3^{1 - n}}} \right)} \] Show SolutionThere really isn’t all that much to this problem. Just remember that, in this case, we’ll need to increase the initial value of the index by two so it will start at \(n = 3\) and this means all the \(n\)’s in the series terms will need to decrease by the same amount (two in this case…).
Doing this gives the following series.
\[\sum\limits_{n = 1}^\infty {\left( {n{2^n} - {3^{1 - n}}} \right)} = \sum\limits_{n = 3}^\infty {\left( {\left( {n - 2} \right){2^{n - 2}} - {3^{1 - \left( {n - 2} \right)}}} \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sum\limits_{n = 3}^\infty {\left( {\left( {n - 2} \right){2^{n - 2}} - {3^{3 - n}}} \right)} }}\]Be careful with parenthesis, exponents, coefficients and negative signs when “shifting” the \(n\)’s in the series terms. When replacing \(n\) with \(n - 2\) make sure to add in parenthesis where needed to preserve coefficients and minus signs.