Mathematical Induction problem

Click For Summary
SUMMARY

The forum discussion centers on proving the formula for the sum of cubes, specifically that \(1^3 + 2^3 + \ldots + n^3 = (1 + 2 + \ldots + n)^2\), using mathematical induction as outlined in Spivak's Calculus. The user initiates the proof by verifying the base case for \(n=1\) and \(n=2\) and attempts to establish the inductive step by expressing the sum for \(n+1\). The discussion emphasizes the importance of correctly expanding the right-hand side and applying known summation formulas to complete the proof.

PREREQUISITES
  • Understanding of mathematical induction principles
  • Familiarity with summation formulas, particularly \(1 + 2 + \ldots + n = \frac{n(n+1)}{2}\)
  • Knowledge of algebraic expansion techniques
  • Basic calculus concepts as presented in Spivak's Calculus
NEXT STEPS
  • Study the properties of mathematical induction in depth
  • Learn how to derive and apply the formula for the sum of the first \(n\) integers
  • Explore algebraic identities and their applications in proofs
  • Practice additional problems involving mathematical induction and summation
USEFUL FOR

Students studying calculus, particularly those self-learning from Spivak's Calculus, and anyone interested in mastering mathematical induction and algebraic proofs.

Fizex
Messages
201
Reaction score
0
I'm trying to solve this problem from CH2 of spivak's calculus of which I am self-studying.

Homework Statement


Prove the following by mathematical induction:
[tex]1^3+...+n^3=(1+...+n)^2[/tex]


Homework Equations


To prove by mathematical induction, you test whether P(1) is true and if P(k) is true then P(k+1) is true.


The Attempt at a Solution


[tex]1^3+...+n^3=(1+...+n)^2[/tex]
[tex]1^3+...+n^3+(n+1)^3=(1+...+n)^2+(n+1)^3[/tex]
[tex](1+...+n)^2+(n+1)^3=(1+...+n+1)^2[/tex]

From there I have no clue and I've been staring at that for 15 minutes.
 
Physics news on Phys.org
Don't forget that the RHS is

[tex](1+2+...+n+(n+1))^2[/tex]

(I'm just making sure that we are on the same page here, since what you wrote can be mistaken as missing that extra n).

Now take the RHS and treat 1+2+...+n as one constant on its own, and expand.
 
Show that it's true for n=1 (or n=2 even)
1^3=1^2, true for n=1, 1^3+2^3=9=(1+2)^2=3^2=2, so true for n=2. Suppose there is a k such that:
[tex] 1^{3}+2^{3}+\cdots +k^{3}=(1+2+\cots +k)^{2}[/tex]
So, we write:
[tex] (1+2+\cdots +k+(k+1))^{2}=(1+2+\cdots +k)^{2}+(k+1)^{2}+2(k+1)(1+2+\cdots +k)[/tex]
Now you should know a formula for (1+2+...+k), the equation becomes:
[tex] (1+2+\cdots +k+(k+1))^{2}=1^{3}+2^{3}+\cdots +k^{3}+(k+1)^{2}+2(k+1)(1+2+\cots +k)[/tex]
So you now have to show that:
[tex] (k+1)^{3}=(k+1)^{2}+2(k+1)(1+2+\cdots +k)[/tex]
How would you go about doing this?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K