Calculate total money saved in the bank Here's the question.
At the beginning of a year, James deposited 2000 dollars in the bank. The annum dividend is 4%. For every subsequent beginning of the year, he will deposit another 260 dollars in the bank. Hence, derive a equation for the amount of money he get at the end of nth year. I need year. It seems like it is impossible to derive a equation. The question is under chapter sequences and series. Thank you!
 A: He starts with $a_0 = 2000$.
Then at the beggining of the next year $n \geq 1$ he will have $(what\ he\ had\ so\ far) \cdot 1.04 + 260$, this means $a_n = 1.04 a_{n-1} + 260$.
$$
a_n = 1.04 a_{n-1} + 260 \\
a_n = 1.04 (1.04 a_{n-2} + 260)  + 260 = 1.04^2 a_{n-2} + 1.04\cdot 260 + 260\\
a_n = 1.04 ( 1.04 (1.04 a_{n-3} + 260) + 260 ) + 260 = 1.04^3 a_{n-3} + 1.04^2 \cdot 260 + 1.04 \cdot 260 + 260
$$
We can conjecture that :
$$
a_n = 1.04^n \cdot 2000 + 260 \cdot \sum_{i=0}^{n-1} 1.04^i = 2000\cdot 1.04^n + 260 \cdot \frac{1.04^n -1}{1.04 -1}
$$
Then use induction to prove the conjecture.
A: HINT....Start at the beginning and develop a pattern.
At the end of the first year, he has $$2000\times 1.04$$
At the end of the second year he has $$(2000\times1.04+260)\times1.04=2000\times1.04^2+260\times 1.04$$
If you continue in this way, and use the sum of a geometric series...
A: Denote by $a_n$ the amount of money banked at the beginning of period $n$ where $2000$ is deposited at $n=0$. Let $r=0.04$ and $K=260$ is deposited at the beginning of each period.
Note that
$$a_{n+1}=(1+r)a_n+K$$
is an expression of how much money you have in the bank one-year-ahead.
Solve the recursion with initial condition $a_0=2000$.
Hint: This is an inhomogenous difference equation of order $1$, so an ansatz will probably do...
A: Let $a_n$ denote the money we deposit in the $n$'th year and $b_n$ the money at the end of the year.
Lets look for some pattern:
First year we deposit $a_0$, so in the end of the year we have:
$b_0 = 1.04 a_0$. 
Next year we see:
$b_1 = 1.04(b_0 + a_1) = 1.04^2 a_0 + 1.04 a_1$
And yet another:
$b_2  = 1.04(b_1 + a_2) = 1.04^3 a_0 + 1.04^2 a_1 + 1.04 a_2$ 
We can now generalize for any n (using induction for example).
$b_n = \sum\limits_{i = 0}^{n} 1.04^{(n-i+1)}a_i$
Plugging the number you gave us:
$b_n = 1.04^{n+1} \cdot 2000 +260 \sum\limits_{i = 1}^{n} 1.04^{(n-i+1)}$
Slightly rearranging:
$b_n = 1.04^{n+1} \cdot 2000 +260 \cdot 1.04^n \sum\limits_{i = 0}^{n-1} \frac{1}{1.04}^{i}$
The last term is known as geometric series and equals:
$b_n = 1.04^{n+1} \cdot 2000 +260 \cdot 1.04^n \frac{1- (\frac{1}{1.04})^n}{1 - \frac{1}{1.04}}$
Now all that's left is some cleaning up:
$b_n = 1.04^{n+1} \cdot 2000 +270.4 \cdot \frac{1.04^n- 1}{0.04}$
