How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?
 A: 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 $n=3$.
A: It's equivalent to $$5\cdot2^n=4n+4+n\cdot2^n$$
Which requires $-1<n<5$. You just have to try $n=0,1,2,3,4$
A: 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 $n=1,n=2,n=3$
A: $\Large \frac{2^{n+1}}{n+1}=\frac{4+2^{n}}{3}$
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 $0,1,2,3,4$
