Closed form solution of a difference equation

The question I am trying to answer is as follows:

1. A smoker has a daily intake of $0.02$mg of nicotine. It is assumed that $1$% of nicotine is disintegrated by the body per day.

a) Set up a difference equation for the amount of nicotine $N_t$ after $t$ days, starting with an initial level of $N_0=0$.

b) Derive a closed form solution for $N_t$.

So my attempt was:

a) $N_{t+1}=0.99N_t+0.02$

b) $N_t= \frac{N_{t+1}-0.02}{0.99}$

I do however feel like the second part is incorrect and it is not as simple as solving the prior equation for $N_t$, assuming my part a) is even correct. The term "closed form solution" is what I don't particularly understand.

Any insight would be appreciated!

Note,

$$N_{t+1}=0.99N_t+0.02$$

$$2=0.99(2)+0.02$$

What happens when you subtract the two? We get,

$$(N_{t+1}-2)=0.99(N_t-2)$$

Now let $A_t=N_{t}-2$ so that,

$$A_{t+1}=0.99A_t$$

$$A_t=A_0(0.99)^{t}$$

$$A_t=-2(0.99)^{t}$$

$$N_t=2-2(0.99)^{t}$$

• Could you explain to me please how you did this step: $A_{t+1}=0.99A_t$ to $A_t=A_0(0.99)^t$ ? – Evan Feb 13 '17 at 17:46
• We multiply by $0.99$ each time to get to the next term from $A_0$, if we multiply more than once we have a power. @N.K Does that clear everything up? – Ahmed S. Attaalla Feb 13 '17 at 17:47
• Ah yes I see now, thank you! :) – Evan Feb 13 '17 at 17:50
• How was it that you were able to sub in $2$ for both $N_{t+1}$ and $N_t$ initially? Like what is the reasoning behind it, and would it work for any number other than $2$? @Ahmed – Evan Feb 13 '17 at 17:53
• I've worked with recurrences before of the form $f(x+1)=af(x)+b$. And I know from experience that the number $L$ which satisfies $L=aL+b$ is useful in that way. And interestingly enough it is often the limit of the sequence. @N.K So what I did was solve $L=0.99L+.02$. The solution to this is $L=2$. Then I can use the trick you saw in the answers. – Ahmed S. Attaalla Feb 13 '17 at 17:57