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1. I'm a bit confused with the following question:

Porve by mathematical induction that sigma r = 1/2n(n+1)

I understand all the working out up until:

When n=k+1

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

^ I'm confused on how ot get from the first line to the second line?

2. Your first line is correct, and the final answer. It is what equals when n=k+1.
3. (Original post by Hopple)
Your first line is correct, and the final answer. It is what equals when n=k+1.

I know the first line is correct but I don't know how to get from that first line to the second line?
4. You don't want the second line, throw it away

What you've done in getting the first line is show that if is true for n=k, then it must also be true for n=k+1, which is your inductive step.
5. (Original post by Hopple)
You don't want the second line, throw it away

What you've done in getting the first line is show that if is true for n=k, then it must also be true for n=k+1, which is your inductive step.
Yes but how did it get from 1/2k (k+1) + (k-1)

to 1/2 (k+1)(k+2)
6. (Original post by Kosovarja94)
Yes but how did it get from 1/2k (k+1) + (k-1)

to 1/2 (k+1)(k+2)
do you mean 1/2k (k+1) + (k+1) ?
7. Take (k+1) out as a factor:

(k+1) ( 1/2k + 1)
= (k+1)*(1/2)(k+2)
8. (Original post by kfkle)
do you mean 1/2k (k+1) + (k+1) ?
Sorry, yes I mean 1/2k (k+1) (k+1)
9. So you're having difficulty with thse two lines?

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

Just stick both terms in the same fraction (over 2) and then factorise the quadratic.

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Updated: December 5, 2010
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