Understanding Induction and Finding Formulas for Summations

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Homework Help Overview

The discussion revolves around mathematical induction and summation formulas, specifically focusing on proving the formula for the sum of cubes and finding a formula for the sum of odd integers.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the sum of odd integers and the sum of the first n integers. Questions arise about the steps in the provided solution and the reasoning behind transforming one expression into another.

Discussion Status

Some participants have offered hints and references to known formulas, while others express confusion regarding specific steps in the solution process. There is an acknowledgment of the connection between the sums, but no consensus on the complete understanding of the transformations involved.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the types of assistance they can provide. The original poster indicates a lack of understanding regarding certain steps in the solution, suggesting that further clarification is needed.

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Homework Statement



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

Homework Equations



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


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:

[itex]\sum^{n}_{i=1}(2i-1)=1+2+3+...+2n-2(1+...+n)<br /> =(2n)(2n+1)/2-n(n+1)[/itex]

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


Thanks
 
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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.
 
We can get the odd integers by first starting with all integers and removing those which are even.
 
Got it, thanks.
 

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