What does a summation mean and how do you compute it? An explanation should be really helpful for me in understanding what
$$\sum_{k=1}^{10}(2^k- 2^{k-1})$$
means and how to evaluate it.
 A: It just means for each $(2^k- 2^{k-1})$ substitute the value of $k$ and add it to the next $(2^k- 2^{k-1})$ with the next value of $k$ substituted in it. $\sum_{k=1}^{10}(2^k- 2^{k-1})$ is shorthand for this and means summation. (do this 10 times). 
So $$\sum_{k=1}^{10}(2^k- 2^{k-1})= (2^1-2^{1-1}) + (2^2-2^{2-1}) + (2^3-2^{3-1})+ (2^4-2^{4-1})+ (2^5-2^{5-1})+ (2^6-2^{6-1})+ (2^7-2^{7-1})+ (2^8-2^{8-1})+ (2^9-2^{9-1})+ (2^{10}-2^{10-1})$$
I wrote this out explicitly so you could get a 'feel' for what was going on. However, you could just note that $$2^k- 2^{k-1}=2^k- 2^{k}\times 2^{-1}=2^k(1-2^{-1})=2^k\left(1-\frac12 \right)=\cfrac{2^k}{2^1}=2^{k-1}$$  
So
$$\sum_{k=1}^{10}(2^k- 2^{k-1})=\sum_{k=1}^{10}2^{k-1}$$ which makes your life easier. You can even see this intuitively by looking at the sum I fully wrote out and observing the cancellation taking place between corresponding terms in the series.
If you have a very large number for $k$ say $k=1000$. Since this series $$1+2+4+8+16+\cdots=2^0+2^1+2^2+2^3+2^4+\cdots$$ is a geometric progression.
You can therefore use the formula for the sum of a geometric series which is $=\cfrac{a(1-r^{k})}{1-k}$. Where $a$ is the first term and $r$ is the common ratio. $r=\cfrac{\mathrm{next}\space \mathrm{term}}{\mathrm{previous}\space \mathrm{term}}=\cfrac84=2$ and $a=1$ 
$$\sum_{k=1}^{1000}2^{k-1}$$
$$=\cfrac{1(1-2^{1000})}{1-2}$$
$$=\cfrac{1-2^{1000}}{-1}$$
$$=2^{1000}-1$$
It will take a computer to actually calculate this so here it is:

A: Hint. In general,
$$
\begin{align}
\sum_{k=1}^{10}(u_k- u_{k-1})&=\sum_{k=1}^{10}u_k-\sum_{k=1}^{10}u_{k-1}\\\\
&=\sum_{k=1}^{10}u_k-\sum_{p=0}^{9}u_{p}\qquad (p=k-1)\\\\
&=u_{10}-u_0 \qquad (\text{many terms cancel}).\\\\
\end{align}
$$
