## Precalculus (6th Edition) Blitzer

$\frac{1}{3}$, -$\frac{1}{5}$ , $\frac{1}{9}$, -$\frac{1}{17}$
To ﬁnd the first four terms of the sequence whose general term is $a_{n}$ = $\frac{(-1)^{n+1}}{2^{n} +1}$, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $\frac{(-1)^{1+1}}{2^{1} +1}$ =$\frac{1}{3}$ n=2, $a_{2}$ =$\frac{(-1)^{2+1}}{2^{2} +1}$ = -$\frac{1}{5}$ n=3, $a_{3}$ = $\frac{(-1)^{3+1}}{2^{3} +1}$ =$\frac{1}{8 +1}$ = $\frac{1}{9}$ n=4,$a_{4}$ = $\frac{(-1)^{4+1}}{2^{4} -1}$ = $\frac{-1}{16 + 1}$ = $\frac{-1}{17}$ The ﬁrst four terms are $\frac{1}{3}$, -$\frac{1}{5}$ , $\frac{1}{9}$, -$\frac{1}{17}$.