What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right).$$ 
How is this possible?
$$\sum_{k=1}^{n+3}\left(2 \times\frac{1}{4}\right)^{k}\left(\frac{1}{4}\right)^n\left(\frac{1}{4}\right)^{-k} \rightarrow \left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3} 2^k\left(\frac{1}{4}\right)^k  \left(\frac{1}{4}\right)^{-k}\rightarrow \left(\frac{1}{4}\right)^n\left(2^{n+3}-1\right)$$
 A: The formula $$\sum_{k=1}^{n}ar^{k}=a\left(\frac{1-r^{n}}{1-r}\right)$$ is incorrect. The correct formula is $$\sum_{{\Large k=}{\Huge0}}^{\Huge n-1}ar^{k}=a\left(\frac{1-r^{n}}{1-r}\right)$$ or $$\sum_{k=1}^{n}ar^{k}=a{\Large{r}}\left(\frac{1-r^{n}}{1-r}\right)$$. With $a=1$ and $r=2$, you should have $\sum_{k=1}^{n+3}1\cdot2^{k}=2\left(\frac{1-2^{n+3}}{1-2}\right)=1\cdot2\left(2^{n+3}-1\right)$
A: Note that 
$$
\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}=\left(\frac{1}{4}\right)^n 2^k
$$
So the sum in question simplifies to
$$
\left(\frac{1}{4}\right)^n\left(\sum_{k=1}^{n+3} 2^k\right)
$$
Now by the geometric series formula
$$
\sum_{k=1}^{n+3} 2^k=2\sum_{k=0}^{n+2} 2^k=2(2^{n+3}-1)
$$
So the total sum is
$$
2\left(\frac{1}{4}\right)^n(2^{n+3}-1)
$$
A: Let $n$=0. The sum is then $\sum_{k=1}^{3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{-k}= \frac{1}{2}\cdot4+\frac{1}{4}\cdot16+\frac{1}{8}\cdot64=2+4+8=14$. This equals $2\left(\frac{1}{4} \right)^{0}\left(2^{0+3}-1\right)=2\cdot 7$, and it does not equal $\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$, which equals 7.
How are you getting that the sum equals $\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$?
