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February 5th, 2012, 08:21 PM   #1
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Mathematical Induction technicality help wanted!

Hey anyone can help me with a little confusion here:

Prove that 0+1+...+n = n(n+1)/2

I did the base case, where P(0) is 0 on the left and right hand sides.
I made the inductive hypothesis: ...+k= k(k+1)/2

Then to show P(k+1) is also true:

0+1+...+k +(k+1) = k+1((k+1)+1)/2
Using the i.h. we plug into this and get:

k(k+1)/2 + (k+1) = k+1(k+2)/2

I left the right hand side alone since that's the desired result we need on the left hand side. However the left hand side I got lost.

I tried distributing:
k^2+k/2 + k+1 but that doesn't equal the right hand side!

Unless I'm overlooking things, or just need to see a different algebraic manipulation here. Thanks in advance to those who help me!
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February 5th, 2012, 08:31 PM   #2
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Re: Mathematical Induction technicality help wanted!

(k^2 + k)/2 + k + 1 = (k^2 + k)/2 + 2(k + 1)/2 = [k^2 + k + 2k + 2]/2 = ... The right hand side.

You need a common denominator of 2 on the LHS in order to combine into one fraction.
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