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# Is this induction valid? watch

1. Un = n^2+3n

Prove, Un is divisible by 2

Un = n^2+3n = 2m
=> 3n = 2m-n^2

Un+1 = (n+1)^2+3(n+1)
=> n^2+5n+4
=> n^2+2n+3n+4
=>n^2+2n+2m-n^2+4
=>2(n+m+2)

I assumed that Un was divisible by 2, so equal to 2m, where m is any integer and then used that to prove that Un+1 is divisible by 2. Is this fine for a FP1 exam?
2. I wouldn't think so. You have simply shown that IF Un is divisible by 2 then so also is Un+1.

Hint:try splitting the n^2 + 3n into two factors
3. The induction step is correct. One may write it slightly shorter without introducing m: n^2+5n+4 = (n^2+3n) + 2(n + 2), and the first number is even by IH. Don't forget the base case.
4. You don't need to equate anything to 2m... just state that it's divisible by two and follow on from there

is divisible by 2. Assume true for so is divisible by 2

since by our assumption, is divisible by 2, and clearly is divisible by 2, we can deduct that if is divisible by 2, then so is .

So we just need to show the first case.

which is divisible by 2, so the statement is true for all n greater than or equal to 1.

Do something similar for n = k-1
5. simpler.

Un = n^2 + 3n = n(n+3)

if n is odd, (n+3) is even and product is even
if n is even, then (n+3) is odd and product is even

Hence Un is even.
6. (Original post by steve10)
simpler.

Un = n^2 + 3n = n(n+3)

if n is odd, (n+3) is even and product is even
if n is even, then (n+3) is odd and product is even

Hence Un is even.
That is a good proof, but it's not proof by induction, so although it's correct it would attract no marks in an exam question about inductive proofs
7. Wicked, thanks guys.

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