I've been reading a PDF of slides from my Discrete Math I professor. The title is Sums, Products and Asymptotic Estimations.
He gives us a problem to fire off the lecture, which is the following:
In a TV game show, the host is willing to give you 50,000€ every year for 20
years, or 750,000€ right in your hands.
What would be the most profitable choice, if you knew that the interest rate
in the bank is 3% every year?
He then goes on, in the notes, and says that the total value of per-year reward is $\sum\limits_{i=0}^{19}\frac{50000}{(i+p)^i}$
Right after that, to calculate the sum, he's got a slide titled "Summation of geometrical series", in which he writes: $\begin{split} S(k) & = x^0 + x^1 + ... + x^k \\ -xS(k) & = -x^1 - x^2 - ... - x^{k+1} \\ S(k) - xS(k) & = 1 - x^{k+1} \Rightarrow S(k) = \frac{1-x^{k+1}}{1-x} \end{split}$
Finally, he finishes by referencing the problem again, this time saying
As we calculated, it is more profitable to get 50,000€ for 20 years
as it has a value of 766,190.
Could anyone point me in the right direction or assist me in finding out how he came to that conclusion?
I cannot, for the life of me, understand how he ended up with that result. Isn't $x$ here supposed to be $\frac{1}{1.03^i}$ ?
The details depend on whether we get the first $50000$ right now, or a year from now. We will assume right now.
We need to calculate the present value (PV) of $50000$ that we get $k$ years from today. This is the amount $A_k$ which if invested would grow to $50000$ in $k$ years.
Now $A_k$, in $k$ years, grows to $A_k(1.03)^k$. If this is $50000$, then $A_k=50000(1.03)^{-k}$. So the present value of the $20$ payments is $$50000+50000(1.03)^{-1}+50000(1.03)^{-2}+\cdots +50000(1.03)^{-19}.$$
To calculate, we need to find the sum of the geometric series $$1+x+x^2+\cdots+x^{19},$$ where $x=(1.03)^{-1}$. To find a formula for the sum, let $S=1+x+x^2+\cdots +x^{19}$. Then $xS=x+x^2+x^3+\cdots +x^{20}$. Subtract. There is a lot of cancellation. We get $$S-xS=(1-x)S=1-x^{20},$$ which gives us $$S=\frac{1-x^{20}}{1-x}.$$ The rest is calculator work. Plug in $(1.03)^{-1}$ for $x$, and multiply by $50000$.