Chapter 9.5, Problem 9QR

Solution

To calculate : the value of the function $P\left(n\right)=\frac{2n}{3n+1}$ for $P\left(1\right)$ , $P\left(k\right)$ and $P\left(k+1\right)$ .

The value of the function $P\left(n\right)=\frac{2n}{3n+1}$ for $P\left(1\right)$ , $P\left(k\right)$ and $P\left(k+1\right)$ are $-2$ , $2k^2-k-3$ and $\left(2k-1\right)\left(k+2\right)$ .

Given information :

The function $P\left(n\right)=\frac{2n}{3n+1}$ .

Formula used :

Substitute the desired value into the function.

Calculation :

Consider the function $P\left(n\right)=\frac{2n}{3n+1}$ ,

Substitute 1 for $n$ into the function $P\left(n\right)=\frac{2n}{3n+1}$ .

  $\begin{array}{c}P\left(1\right)=\frac{2\left(1\right)}{3\left(1\right)+1}\\ =\frac{2}{3+1}\\ =\frac{2}{4}\\ =\frac{1}{2}\end{array}$

Substitute $k$ for $n$ into the function $P\left(n\right)=\frac{2n}{3n+1}$ .

  $P\left(k\right)=\frac{2k}{3k+1}$

Substitute $k+1$ for $n$ into the function $P\left(n\right)=\frac{2n}{3n+1}$ .

  $\begin{array}{c}P\left(k+1\right)=\frac{2\left(k+1\right)}{3\left(k+1\right)+1}\\ =\frac{2k+2}{3k+3+1}\\ =\frac{2k+2}{3k+4}\end{array}$

Factorize the polynomial $2k^2+3k-2$ .

  $\begin{array}{c}P\left(k+1\right)=2k^2+4k-k-2\\ =2k\left(k+2\right)-1\left(k+2\right)\\ =\left(2k-1\right)\left(k+2\right)\end{array}$

Thus, the value of the function $P\left(n\right)=\frac{2n}{3n+1}$ for $P\left(1\right)$ , $P\left(k\right)$ and $P\left(k+1\right)$ are $\frac{1}{2}$ , $\frac{2k}{3k+1}$ and $\frac{2k+2}{3k+4}$ .