Question

find P(k+1) if P(k) = 2^(k-1) / k! (the exclamation mark is not me getting excited. i don't know what it's there for)

Answer 1

"i don't know what it's there for"


`!` is the factorial symbol, and it's used to denote the product of successively smaller non-negative integers. That is, if
`n=10`, for instance, then
`n!` is the product of all positive integers less than or equal to 10:


`10!=10\cdot9\cdot8\cdot\ldots\cdot3\cdot2\cdot1`

By convention, we define
`0!=1`.

One way of interpreting the factorial is combinatorically, meaning we can use it to count the number of ways of doing some particular task. Suppose I'm getting dressed, and I have 3 basic items that I have to put on: a shirt, a pair of pants, and a pair of socks. How many ways are there in which I can put each thing on in order?

For the first item I have 3 choices (shirt, pants, or socks). Once I put on the first article of clothing, I have 2 choices left. After that, 1 choice is left. In total, I would have
`3\cdot2\cdot1=6` possible ways of putting on all my clothes in a particular order.

The factorial generalizes this and makes writing the product more compact:
`3!=3\cdot2\cdot1=6`.

Back to your actual question: We're given some function
`P(k)`, and we're asked to find the value of
`P(k+1)`. This is just a matter of replacing
`k` with
`k+1`:


`P(k+1)=(2^((k+1)-1))/((k+1)!)=(2^k)/((k+1)!)`

and we're done.