Calculating a cumulative series I could do this on a calculator but I thought there would be a more efficient way to do this than punching on my calculator the following:

$3.5 \cdot 1.03 + 3.5 \cdot 1.03^2 + 3.5 \cdot 1.03^3 + \ldots + 3.5 \cdot 1.03^{10}$

So basically,

$xy + xy^2 + xy^3 + ... + xy^{10}$

How can I simplify this to make it easier to calculate on a calculator?
 A: This is called a finite geometric series. It has a formula. Let's derive it.
Let $$S = a + a(r)+ a(r)^2 + \dots + a(r)^n.$$ 
With this set up the quantity you seek is $S - a$, since you don't want the first term. Let's get a formula for $S$ and then just subtract $a$.
Multiplying $S$ and $r$ gives $$Sr = a(r) + \dots + a(r)^{n+1}.$$
Then the difference $$Sr - S = a(r)^{n+1} - a.$$ If that step is not clear, write it out term by term and do the subtraction to see everything cancel except for what is on the right hand side above. Now factor to get $$S(r-1) = a(r)^{n+1} - a$$ which we can divide to to yield $$S = \frac{a(r)^{n+1} - a}{r-1}.$$ So now subtract $a$ to get the value you want, $$S - a = \frac{a(r)^{n+1} - a}{r-1} - a = a \cdot \frac{r^{n+1} - r}{r-1}  .$$
I did a lot of simplification in the last equality, but I am sure you can work that out on your own. To solve your explicit problem, plugin and evaluate with r = 1.03, a = 3.5, and n=10 to get $$(3.5) \cdot \frac{(1.03)^{11} - 1.03}{(.03)} \approx 41.327.$$ Good question by the way, it is always good to be curious :)
