Hello, helloying!
There must be typos in the problem.
The claim is not true.
.[1]
. . . . . . . .
Prove by induction that: . .[2]
The recursion formula has a square root.
I don't see how the term could always be an integer.
In fact, we can't determine the second term of the sequence!
In [1], let
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
In [2], let
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
1. Write the claim that you need to prove in the following form: "for all n >=1, ... ". Call the part denoted by ellipses, which contains n, by P(n). This is the induction statement, or the induction hypothesis (IH). Note that for each particular n, P(n) is not a number; it is a proposition that is either true or false.
2. Prove P(1). This is the base case.
3. Prove "for all n, P(n) implies P(n+1)". For this, fix an arbitrary n and assume P(n) (that's why it is called the IH). Use it to prove P(n+1).