Spivak 2-1b, 2-2a (induction)

[carlosbgois]

1. The problem statement, all variables and given/known data

1b) Prove by induction: 1^{3}+...+n^{3}=(1+...+n)^{2}
2a) Find a formula for: \sum^{n}_{i=1}(2i-1)

2. Relevant equations

There's a Hint for 2a): 'What to this expression have to do with 1+2+3+...+2n?'


3. The attempt at a solution

In 2a) I've got near the answer, when comparing with the given one, but I can't understand the last thing he does. The solution in the book is:

\sum^{n}_{i=1}(2i-1)=1+2+3+...+2n-2(1+...+n)
=(2n)(2n+1)/2-n(n+1)

And I couldn't understand how to make the second member become the third one, which goes directly to the answer n^{2}


Thanks

[gb7nash]

It's a well known fact that for positive integer n:

1+2+3+...+n = n(n+1)/2

Use this to obtain the answer.

[206PiruBlood]

We can get the odd integers by first starting with all integers and removing those which are even.

[carlosbgois]

Got it, thanks.