Mathematical Induction problem

AI Thread Summary
The discussion revolves around proving the equation 1^3 + ... + n^3 = (1 + ... + n)^2 using mathematical induction. The initial steps involve verifying the base case for n=1 and n=2, confirming the formula holds true for these values. The next step requires expressing the induction hypothesis for k and then expanding the right-hand side to include (k+1). The key challenge is to demonstrate that (k+1)^3 equals (k+1)^2 plus 2(k+1)(1 + 2 + ... + k). The conversation emphasizes the importance of correctly applying the formula for the sum of the first k integers in this proof.
Fizex
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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:
1^3+...+n^3=(1+...+n)^2


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


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

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

(1+2+...+n+(n+1))^2

(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:
<br /> 1^{3}+2^{3}+\cdots +k^{3}=(1+2+\cots +k)^{2}<br />
So, we write:
<br /> (1+2+\cdots +k+(k+1))^{2}=(1+2+\cdots +k)^{2}+(k+1)^{2}+2(k+1)(1+2+\cdots +k)<br />
Now you should know a formula for (1+2+...+k), the equation becomes:
<br /> (1+2+\cdots +k+(k+1))^{2}=1^{3}+2^{3}+\cdots +k^{3}+(k+1)^{2}+2(k+1)(1+2+\cots +k)<br />
So you now have to show that:
<br /> (k+1)^{3}=(k+1)^{2}+2(k+1)(1+2+\cdots +k)<br />
How would you go about doing this?
 
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