How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$

Question

$n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?

Answer 1

Rewriting, we have $3\cdot2^{n+1}=(n+1)(4+2^n)$

First, note that whenever $n\geq 5$, we have $(n+1)(4+2^n)>6\cdot 2^n=3\cdot2^{n+1}$

Next note that whenever $n<-1$, $(n+1)(4+2^n)<0<3\cdot2^{n+1}$

Note that $n=-1$ isn't possible because in the first equation we would divide by $0$.

Hence any solution must have $n\in[0,4]$

Now we just check our $5$ cases and find that the only solutions are $n=1$, $n=2$, and $n=3$.

Answer 2

It's equivalent to $$5\cdot2^n=4n+4+n\cdot2^n$$ Which requires $-1<n<5$. You just have to try $n=0,1,2,3,4$

Answer 3

the equation $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ is equivalent to $$2^{n-2}=\frac{n+1}{5-n}$$

because $2^{n-2}>0$ follow that $\frac{n+1}{5-n}>0,0<n<5$

we need to check for $n=1,2,3,4$

for $n=1$ we get $$2^{-1}=\frac{2}{4}=\frac{1}{2}$$ for $n=2$ we get $$2^0=\frac{3}{3}=1$$ for $n=3$ we get $$2^1=\frac{4}{2}=2$$ for $n=4$ we get $$2^2\neq\frac{5}{1}=5$$ so the integer solutions are $n=1,n=2,n=3$

Answer 4

$\Large \frac{2^{n+1}}{n+1}=\frac{4+2^{n}}{3}$

this equation tells us that $n+1>0 \Rightarrow n>-1 $((i)we can't divide by zero (ii) RHS is +ve so LHS must also be +ve)

$\Large \frac{2^{n+1}}{n+1}=\frac{8+2^{n+1}}{6}=\frac{8}{6}+\frac{2^{n+1}}{6}$(multiply and divide RHS by $2$)

$\Large \frac{2^{n+1}}{n+1}-\frac{2^{n+1}}{6}=\frac{8}{6}$(difference of two terms is +ve)

this equation tells us that $\Large \frac{2^{n+1}}{n+1} > \frac{2^{n+1}}{6}\Rightarrow n+1<6 \Rightarrow n<5$

So the integer values to check for the answer are $0,1,2,3,4$