How to calculate Net Present Value? So I have this question :
Your boss asked you to evaluate a project with an infinite life. 
Sales and costs project to$ \$1,000 $ and $\$500$ per year, respectively. (Assume sales and costs occur at the end of the year, i.e., profit of $\$500$ at the end of year one.) 
There   is no depreciation and the tax rate is $30 \% $. The real required rate of return is $10\%$. The inflation rate is $4\%$ and is expected to be $4 \%$ forever. Sales and costs will increase at the rate of inflation. If the project costs $\$3,000$, what is the NPV?


*

*$\$500.00$

*$\$1629.62$

*$\$365.38$

*$\$472.22$
On the answer sheet it states that 
$$NPV = -3000+ \cfrac{(1000-50)(1-0.30)}{0.10}$$,
which will give me a result of $ \$500$.
The only thing i do not understand is why is it divided by the real rate of return $0.10$, and not $1+r$ real?
I have done various exercises where I always divide by $1+r$.
Somebody please explain ! Thanks!
 A: The present value of the series of cash flows is as follows;  
$$-3000+\frac{(1000-500)\cdot (1-0.3)}{1.1^1}+\frac{(1000-500)\cdot (1-0.3)}{1.1^2}+\frac{(1000-500)\cdot (1-0.3)}{1.1^3}+\frac{(1000-500)\cdot (1-0.3)}{1.1^4}+\ldots$$
$$=-3000+\sum_{k=1}^{\infty}\frac{(1000-500)\cdot (1-0.3)}{1.1^k}$$
For simplicity let $C=(1000-500)\cdot (1-0.3)$. The infinite sum is $\sum\limits_{k=1}^{\infty}\frac{C}{1.1^k}$
We can look at the partial sum of the geometric series $$\sum\limits_{k=1}^{n}\frac{C}{1.1^k}=C\cdot \frac{1}{1.1}\cdot  \frac{1-\left(\frac{1}{1.1} \right)^n}{1-\frac{1}{1.1}}$$ 
Now we can expand the term by $1.1$ (blue terms)
$$=C\cdot \frac{\color{blue}{1.1}}{1.1}\cdot  \frac{ 1-\left(\frac{1}{1.1} \right)^n}{\color{blue}{1.1}\cdot \left(1-\frac{1}{1.1}\right)}=C\cdot  \frac{ 1-\left(\frac{1}{1.1} \right)^n}{1.1-1}=C\cdot  \frac{ 1-\left(\frac{1}{1.1} \right)^n}{0.1}$$
Finally $n$ goes to infinty. $\left(\frac{1}{1.1} \right)$ is smaller than $1$. It decreases when n increases. Therefore
$$\lim_{n \to \infty} C\cdot  \frac{ 1-\left(\frac{1}{1.1} \right)^n}{0.1}= C\cdot  \frac{ 1-0}{0.1}=\frac{C}{0.1}$$
I think it is clear from where the $0.1$ comes.
