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    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?
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    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
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    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.
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    You don't need to equate anything to 2m... just state that it's divisible by two and follow on from there

    u_n = n^2+3n

    u_n is divisible by 2. Assume true for n=k so u_k=k^2+3k is divisible by 2

    u_{k+1} = (k+1)^2+3(k+1) = k^2+5k+4 = (k^2+3k)+(2k+4)

    since by our assumption, (k^2+3k) is divisible by 2, and clearly (2k+4) is divisible by 2, we can deduct that if u_k is divisible by 2, then so is u_{k+1}.

    So we just need to show the first case.

    u_1 = 4 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
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    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.
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    (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
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    Wicked, thanks guys.
 
 
 
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