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Section 10.1 : Sequences
1. List the first 5 terms of the following sequence.
\[\left\{ {\frac{{4n}}{{{n^2} - 7}}} \right\}_{n = 0}^\infty \] Show SolutionThere really isn’t all that much to this problem. All we need to do is, starting at \(n = 0\), plug in the first five values of \(n\) into the formula for the sequence terms. Doing that gives,
\[\begin{align*} & n = 0:\hspace{0.5in}\frac{{4\left( 0 \right)}}{{{{\left( 0 \right)}^2} - 7}} = 0\\ & n = 1:\hspace{0.5in}\frac{{4\left( 1 \right)}}{{{{\left( 1 \right)}^2} - 7}} = \frac{4}{{ - 6}} = - \frac{2}{3}\\ & n = 2:\hspace{0.5in}\frac{{4\left( 2 \right)}}{{{{\left( 2 \right)}^2} - 7}} = \frac{8}{{ - 3}} = - \frac{8}{3}\\ & n = 3:\hspace{0.5in}\frac{{4\left( 3 \right)}}{{{{\left( 3 \right)}^2} - 7}} = \frac{{12}}{2} = 6\\ & n = 4:\hspace{0.5in}\frac{{4\left( 4 \right)}}{{{{\left( 4 \right)}^2} - 7}} = \frac{{16}}{9}\end{align*}\]So, the first five terms of the sequence are,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left\{ {0,\,\, - \frac{2}{3},\,\, - \frac{8}{3},\,\,6,\,\,\frac{{16}}{9},\,\, \ldots } \right\}}}\]Note that we put the formal answer inside the braces to make sure that we don’t forget that we are dealing with a sequence and we made sure and included the “…” at the end to reminder ourselves that there are more terms to this sequence that just the five that we listed out here.