# Mathematical induction (sum of the first few even numbers)

So the question was basically " Suppose that there are n teams in a rugby league competition. Every team A plays every other team B twice, once at the home ground for team A, and the other time at the home ground for team B."

2(n 1) + 2(n 2) + 2(n 3) + : : : + 6 + 4 + 2 is given

a) Write the expression in summation notation. b) Use mathematical induction to prove it, n>=2

So I got this expression for (a) n^Sigma(i=1) = (2(n-i)) where n is the number of teams

Part B

Proof:

Let P(n) denote the sequence n^Sigma(i=1)=2(n-i) and n≥2

Consider P (2) n^Sigma(i=1)=2(n-i) =2(2-1)=2 ∴it is true when n=2

We will now assume it is true for P(k)

k^Sigma(i=1)=2(k-i) for some integer k ≥2

Consider P(k+1)

k+1^Sigma(i=1)=2(k+1-i) for some integer k ≥2

P(k+1)=2(k-1)+2(k-2)+2(k-3)+⋯+2(k-i)+2(k+1-1)

Since we have assumed that P(k) is true.

So we know: P(k+1)=P(k)+(k+1)

ANSWER i cant answer my own question for 8hrs so here it is:

P(n)=n^2-n

P(K+1)=P(k)+2((k+1)-1)

P(K+1)=〖(k〗^2-k)+2(k+1)-2

P(K+1)=〖(k〗^2-k)+2(k+1)-2

P(K+1)=〖(k〗^2-k)+2k+2-2

P(K+1)=〖(k〗^2-k)+2k

P(K+1)=〖(k〗^2+k)

P(K+1)=(k+1)^2-(k+1)

Therefore under induction the sequence has been proven.

Thanks to @P..

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The answer for part (a) is correct. But for part (b) what are you trying to prove? What do you think $P(k)$ equals to? Hint: Arithmetic progression. – P.. Apr 24 '13 at 6:16
well P(k)=2(k-i)=2k-2i. Know P(K+1)=P(k)+k+1= 2(k-i)+(k+1) which brings me right back to where i am. – Ghozt Apr 24 '13 at 6:19
So can you tell me from your expression what $P(3)$, or $P(5)$ is? The answer should be a number. – P.. Apr 24 '13 at 6:20
@P.. So is P(k)=K^2-k? – Ghozt Apr 24 '13 at 7:18
Yes! Now try to prove it using induction. If you need any help let me know. – P.. Apr 24 '13 at 7:21

The OP edited in an answer to his/her post; I'm copying it here so the question isn't "unanswered."

$P(n)=n^2-n$

$P(k+1)=P(k)+2((k+1)-1)$

$P(k+1)= (k^2-k)+2(k+1)-2$

$P(k+1)=(k^2-k)+2(k+1)-2$

$P(k+1)=(k^2-k)+2k+2-2$

$P(k+1)=(k^2-k)+2k$

$P(k+1)=(k^2+k)$

$P(k+1)=(k+1)^2-(k+1)$

Therefore under induction the sequence has been proven.

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