Answer
a) $k_n = 2\times k_{n-1} + k_{n-5}$ , for $n\gt5$
b) $k_0 = 1, k_1 = 2, k_2 = 4, k_3 = 8, k_4 = 16$
c) $k_{10} = 1217 $
Work Step by Step
Let $k_n$ be the number of ways to deposit $n$ dollars.
We can deposit a dollar bill and then deposit the remaining $n-1$ dollars in $k_{n-1}$ ways.
Or we could also have deposited a dollar coin and then deposit the remaining $n-1$ dollars in $k_{n-1}$ ways.
The last alternative, which only works for $n\geq5$ is to deposit a 5 dollar bill and then deposit the remaining $n-5$ dollars in $k_{n-5}$ ways.
Therefore, $a_n = a_{n-1} + a_{n-1}+a_{n-5}$
Or simply $a_n = 2\times a_{n-1} + a_{n-5}$
b)
The initial conditions are the values of $a_0, a_1, a_2, a_3$ and $a_4$ .
All of the rest values are determined by these.
Clearly, $a_0 = 1$ (only one way to not deposit anything)
$a_1 = 2$ (The bill or the coin)
$a_2 = 2^2 = 4$ (Two choices each with two options)
Similarly, $a_3 = 2^3 = 8$ and $a_4 = 2^4 = 16$
c)
Recursively calculationg each of the $a_5, a_6 . . . a_{!0}$ gives,
$a_5 = 2a_4 + a_0$ = 2 · 16 + 1 = 33
$a_6 = 2a_5 + a_1$ = 2 · 33 + 2 = 68
$a_7 = 2a_6 + a_2$ = 2 · 68 + 4 = 140
$a_8 = 2a_7 + a_3$ = 2 · 140 + 8 = 288
$a_9 = 2a_8 + a_4$ = 2 · 288 + 16 = 592
$a_{10} = 2a_9 + a_5$ = 2 · 592 + 33 = 1217
Thus $a_{!0} = 1217$ is the number of ways to deposit for a book of stamps costing $10.
