How do I solve the following recursive equation?

Question

$$\begin{cases} p(1) = \frac{1}{100} \\ p(t) = p(t-1)\left(1+\frac{1}{101-t}\right) \end{cases}$$

I feel like I could use the Telescoping Technique, but I do not quite see how.

Answer 1

Write as $$(101-t)p(t)=(101-(t-1))p(t-1).$$

As this works for all $t$, the two members are a constant, equal to

$$(101-1)\frac1{100}.$$

Answer 2

Hint: Write it as a product and use $1+\frac{1}{101-n}=\frac{102-n}{101-n}$, then you can apply the telescoping technique.

Answer 3

$$p(t)=\frac{102-t}{101-t}\, p(t-1)$$ for $t=1$ initial condition is $p(1)=\frac{1}{100}=\frac{1}{101-1}$

for $t=2$ we have $p(2)=\frac{100}{99}\,p(1)=\frac{100}{99}\,\frac{1}{100}=\frac{1}{99}=\frac{1}{101-2}$

for $t=3$, $p(3)=\frac{99}{98}\,p(2)=\frac{100}{99}\,\frac{99}{98}\frac{1}{100}=\frac{1}{98}=\frac{1}{101-3}$

for general $t$ we have $$p(t)=\frac{1}{101-t}$$