Question

In Exercises 31–36 provide the first five terms of the given sequence. Unless specified, assume that the first term has index 1.

$\left\{\frac{(-1{)}^{k-1}{x}^{2k}}{2k!}\right\}$

Short answer

The first five terms are$\frac{x^2}{2!},-\frac{x^4}{4!},\frac{x^6}{6!},-\frac{x^8}{8!},\frac{x^{10}}{10!}$.

Step-by-step solution

Step 1. Given information.

The given sequence is$\left\{\frac{(-1{)}^{k-1}{x}^{2k}}{2k!}\right\}$.

Step 2. First term.

The general sequence is $a_k=\frac{(-1{)}^{k-1}{x}^{2k}}{2k!}$.

When $k=1,$

$$\begin{gathered} a_1=\frac{(-1{)}^{1-1}{x}^{2\left(1\right)}}{2\left(1\right)!} \\ =\frac{x^2}{2!} \end{gathered}$$

Step 3. Remaining terms.

Now, put

$$\begin{gathered} k=2, \\ {a}_2=\frac{(-1{)}^{2-1}{x}^{2\left(2\right)}}{2\left(2\right)!} \\ {a}_2=\frac{-x^4}{4!} \\ k=3, \\ {a}_3=\frac{(-1{)}^{3-1}{x}^{2\left(3\right)}}{2\left(3\right)!}=\frac{x^6}{6!} \\ k=4, \\ {a}_4=\frac{(-1{)}^{4-1}{x}^{2\left(4\right)}}{2\left(4\right)!}=\frac{-x^8}{8!} \\ k=5, \\ {a}_5=\frac{(-1{)}^{5-1}{x}^{2\left(5\right)}}{2\left(5\right)!}=\frac{x^{10}}{10!} \end{gathered}$$