# Use mathematical induction to prove the statements are correct for n `in` Z+(set of positive integers). 1) Prove that for n `>= 1` `1^3 + 2^3 +3^3+...+n^3=[n(n+1)]^2/4 `

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According to the method of mathematical induction, the given statement is true if the following conditions are met:

1) The statement is true for n = 1

2) If it is assumed that the statement is true for n, it can be proven that the statement will be true for n+ 1 as well.

It is easy to see that the given statement is true for n = 1:

`1^3 = 1`

`[n(n+1)]^2/4 = [1(1+1)]^2/4 = 1`

It can also be shown, just to illustrate, that the statement will be true for n = 2 as well:

`1^3+2^3 = 9`

`[n(n+1)]^2/4 = [2(2+1)]^2/4 = 9`

Now let's prove that if the statement is true for n, that is, if

`1^3 + 2^3 + ... + n^3 = [n(n+1)]^2/4`

then it is also true for n+1, that is,

`1^3 + 2^3 + ...+n^3 + (n+1)^3 = [(n+1)(n+2)]^2/4`

So we have to show that for any n,

`[n(n+1)]^2/4 + (n+1)^3 = [(n+1)(n+2)]^2/4`

Let's work with the left side:

`(n+1)^2 = n^2 + 2n + 1`

and `(n+1)^3 = n^3 + 3n^2 +3n + 1`

Then, left side equals

`(n^2(n^2 + 2n + 1))/4 + (n^3 + 3n^2 + 3n + 1) = (n^4 + 2n^3 + n^2 + 4n^3 + 12n^2 + 12n + 4)/4 = (n^4 + 6n^3 + 13n^2 + 12n + 4)/4`

Now let's open parenthesis on the ride side and show that it equals the same thing:

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

`(n^2 + 3n + 2)^2 = (n^2 + 3n + 2)(n^2 + 3n + 2) = n^4 + 3n^3 + 2n^2 + 3n^3 +9n^2 + 6n+2n^2 + 6n+4 = n^4 + 6n^3 + 13n^2+ 12n + 4`

The numerator of the fraction on the right side equals to the numerator of the fraction on the left side, and both denominators are 4. So the sum of the cubes of integers up to n+ 1 equals the sum of integers up to n, plus the cube of n+ 1.

We have shown that the statement is true for n + 1, assuming that it is true for n. The second step of the induction method works, so the given statement is proven true using mathematical induction.