Sums of Geometric Series Induction

Question

Suppose that today you put 1000 dollars in a deposit account that pays you 1% interest every month. After N months, the account balance will be $ 1000 \times 1.01^N $ dollars. Now consider a variant of this scenario where, beginning one month from today, you make a deposit of 100 dollars each month. So, for example, after 1 month your account will have $ 1000 \times 1.01 + 100 = 1210 $ dollars; note that the interest is applied to the funds that have been sitting into the account for the past month, and not to the new deposit. Find an expression, using the summation notation, for the account balance after N months under this new scenario. Then use the result of part (a) to find a simpler formula for the balance. What is the balance one year from today?

I understand the sum of the geometric series. However, I do not understand how to prove this through induction. I need help. Thank you.

Answer 1

So basically, what you are looking for is the solution for $$u_{n+1}=1.01u_n+100$$ you should look for : $$v_n=u_n+10000$$ and prove that $(v_n)$ is geometric.

Final answer should be : $$u_n=1.01^n11000 - 100000$$

Answer 2

r.t.p. $$\sum_{i=1}^N r^{i-1}=\frac{1-r^{N}}{1-r} $$

the proposition is true for $N=1$ ( l.h.s. = r.h.s. = 1 )

assume it is true for $N=k$ and prove that it must then be true for $N=k+1$ $$ \begin{eqnarray*} &\sum_{i=1}^{k+1} r^{i-1} \\ =&r^k+\sum_{i=1}^{k} r^{i-1} \\ =&r^k+\frac{1-r^{k}}{1-r} \\ =&\frac{(r^k-r^{k+1}) + (1-r^{k})}{1-r} \\ =&\frac{1-r^{k+1}}{1-r} \end{eqnarray*} $$