To prove that 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2 by mathematical induction, we need to show that the equation holds for the base case (n=1) and then prove that if it holds for n=k, it also holds for n=k+1.
Base case: n = 1
1^3 = (1)^2
This is true, so the equation holds for the base case.
Assume the equation holds for n=k:
1^3 + 2^3 + 3^3 + ... + k^3 = (1 + 2 + 3 + ... + k)^2
We want to prove that the equation holds for n=k+1:
1^3 + 2^3 + 3^3 + ... + (k+1)^3 = (1 + 2 + 3 + ... + (k+1))^2
We can rewrite the left side of the equation as:
1^3 + 2^3 + 3^3 + ... + k^3 + (k+1)^3
Using the assumption for n=k, we can substitute (1 + 2 + 3 + ... + k)^2 for 1^3 + 2^3 + 3^3 + ... + k^3:
(1 + 2 + 3 + ... + k)^2 + (k+1)^3
Expanding the square on the left side, we get:
(1^2 + 2^2 + 3^2 + ... + k^2) + 2(1×2 + 1×3 + ... + (k-1)×k) + (k+1)^3
We can simplify the middle term using the formula for the sum of the first k integers:
1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6
1×2 + 1×3 + ... + (k-1)×k = k(k+1)/2
Substituting these values, we get:
k(k+1)(2k+1)/6 + k(k+1) + (k+1)^3
Simplifying the expression, we get:
(k+1)(k^2 + 3k + 3)/3
Using the formula for the sum of the first k+1 integers, we can rewrite the right side of the equation as:
(1 + 2 + 3 + ... + k + (k+1))^2
Simplifying the expression, we get:
(k+1)(k+2)(2k+3)/6
Now we can see that the left side of the equation is equal to the right side of the equation, so the equation holds for n=k+1.
Therefore, by mathematical induction, the equation 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2 is true for all positive integers n.