Answer
9494 ways.
Work Step by Step
We found in the problem 4 that,
$a_n = a_{n-1}+ a_{n-2} + 2 \times a_{n-5} + 2 \times a_{n-10} + a_{n-20} + a_{n-50} + a_{n-100}$, for $n\geq100$
for all n such that $n\lt0$, we have $a_n = 0 $ since for negative indices we have no ways.
Also, $a_0 = 1$ as there is only one way to pay no money that is not to pay anything.
Therefore, $a _1 = a_0 + negative$ $indices$
or , $a_1 = 1$
Similarly, $a_2 = a_1+a_0 + negative$ $indices$ = 2
$a_3 = a_2+a_1 + negative$ $indices$ = 2 + 1 = 3
$a_4 = a_3+a_2 + negative$ $indices$ = 3 + 2 = 5
$a_5 = a_4+a_3 +2 a_0+negative$ $indices$ = $5 + 3 + 2\times 1$ = 10
We will be dropping negative indices henceforth as it is understood.
$a_6 = a_5 + a_4 + 2a_1$ = 10 + 5 + 2 · 1 = 17
$a_7 = a_6 + a_5 + 2a_2$=17+10 + 2 · 2 = 31
$a_8 = a_7 + a_6 + 2a_3$ = 31 + 17 + 2 · 3 = 54
$a_9 = a_8 + a_7 + 2a_4$ = 54 + 31 + 2 · 5 = 95
Hereafter, the term $a_{n-10}$ will also.start affecting our answer.
$a_{10} = a_9 + a_8 + 2a_5 + 2a_0$ = 95 + 54 + 2 · 10 + 2 · 1 = 171
$a_{11} = a_{10} + a_9 + 2a_6 + 2a_1$ = 171 + 95 + 2 · 17 + 2 · 1 = 302
$a_{12} = a_{11} +a_{10} + 2a_7 + 2a_2$ = 302 + 171 + 2 · 31 + 2 · 2 = 539
$a_{13} = a_{12} + a_{11} + 2a_{8} + 2a_3$ = 539 + 302 + 2 · 54 + 2 · 3 = 955
$a_{14 }= a_{13} + a_{12} + 2a_9 + 2a_4$ = 955 + 539 + 2 · 95 + 2 · 5 = 1694
$a_{15 }= a_{14} + a_{13} + 2a_{10} + 2a_5 $= 1694 + 955 + 2 · 171 + 2 · 10 = 3011
$a_{16} = a_{15} + a_{14} + 2a_{11} + 2a_6 $= 3011 + 1694 + 2 · 302 + 2 · 17 = 5343
$a_{17} = a_{16} + a_{!5} + 2a_{12} + 2a_7$ = 5343 + 3011+2·539+2 · 31 = 9494
Thus, there are $9494$ ways to pay a bill of 17 pesos using
the currency described in Exercise 4.
