Question

In Exercises, find a plausible formula for the general term of the given sequence.

$\left\{\frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9},\dots \right\}$

Short answer

The plausible formula for the general term of the sequence is$a_k=\frac{k+1}{2k+1}$where$k\ge 1.$

Step-by-step solution

Step 1. Given information.

The given sequence is$\left\{\frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9},\dots \right\}.$

Step 2. plausible formula for the general term. 

write the sequence in a pattern.

$$\begin{gathered} \left\{\frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9},\dots \right\}=\left\{\frac{1+1}{2\left(1\right)+1},\frac{2+1}{2\left(2\right)+1},\frac{3+1}{2\left(3\right)+1},\frac{4+1}{2\left(4\right)+1},\dots \right\} \\ \left\{\frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9},\dots \right\}={\left\{\frac{k+1}{2k+1}\right\}}_{k=1}^{\infty } \end{gathered}$$

for every term, the numerator is $\left(k+1\right)$and the denominator is $\left(2k+1\right)$andlocalid="1649189107593" $k\ge 1.$

So plausible formula for the general term of the sequence islocalid="1649189440639" $a_k=\frac{k+1}{2k+1}.$