Understanding Induction and Finding Formulas for Summations

  • Thread starter Thread starter carlosbgois
  • Start date Start date
  • Tags Tags
    Induction Spivak
AI Thread Summary
The discussion focuses on proving by induction that the sum of cubes, 1^3 + ... + n^3, equals (1 + ... + n)^2. Participants also explore finding a formula for the summation of odd integers, expressed as ∑(2i-1) from i=1 to n. A hint suggests relating this summation to the series of integers 1 + 2 + 3 + ... + 2n. The solution involves recognizing that the sum of the first n odd integers equals n^2, derived from manipulating known summation formulas. Understanding these relationships is key to solving both problems effectively.
carlosbgois
Messages
66
Reaction score
0

Homework Statement



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

Homework Equations



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


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)<br /> =(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
 
Physics news on Phys.org
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.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top