Hi. if you find any fault in this proof please let me know since I couldn't find an answer on the back of my math book.

Prove that

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

Let \(\displaystyle P(n)\) be the statement \(\displaystyle \displaystyle\sum_{i=0}^{n} i^3 =1^3 + 2^3 + ... + n^3 = (1+2+3+...+n)^2\)

The sum of the first \(\displaystyle n\) positive integers is \(\displaystyle \frac{n(n+1)}{2}\) which is a theorem refered to as 1-1.

We can rewrite the statement \(\displaystyle P(n)\) based on theorem 1-1 as \(\displaystyle \displaystyle\sum_{i=0}^{n} i^3 = (\frac{n(n+1)}{2})^2\).

Now we will show that \(\displaystyle P(n)\) is true for all integers \(\displaystyle n \geq 0\)

\(\displaystyle P(0)\) is the statement \(\displaystyle \displaystyle\sum_{i=0}^{0} i^3 = (\frac{0(0+1)}{2})^2= 0\) which is clearly true.

Let \(\displaystyle k \geq 0\) be an integer. Assume (for induction) that \(\displaystyle P(k)\) is true. That means \(\displaystyle \displaystyle\sum_{i=0}^{k} i^3\) \(\displaystyle \stackrel{?}{=}\)\(\displaystyle (\frac{k(k+1)}{2})^2\), is true for \(\displaystyle n=k\).

We will prove that \(\displaystyle P(k+1)\) is true as well. That is, we must prove that \(\displaystyle \displaystyle\sum_{i=0}^{k} i^3 + (k+1)^3= (\frac{(k+1)((k+1)+1)}{2})^2\). To prove this equation, start by adding \(\displaystyle (k+1)^3\)

to both sides of the inductive hypothesis:

\(\displaystyle \displaystyle\sum_{i=0}^{k+1} i^3= (\frac{k(k+1)}{2})^2 + (k+1)^3\). Now, simplifying the right side we get:

\(\displaystyle =\frac{k^2((k+1)^2)}{4} + (k+1)^3 = \frac{k^2(k+1^2) +(4k+4)(k+1)^2}{4} = \frac{(k+2)^2(k+1)^2}{2^2} = (\frac{((k+2)(k+1)=}{2})^2 = (\frac{(k+1)((k+1)+1)}{2})^2 \)

Thus \(\displaystyle P(k+1)\) is true, so by the principle of mathematical induction \(\displaystyle P(n)\) is true for all integers \(\displaystyle n \geq 0\)

Q.E.D.

Prove that

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

**Statement**Let \(\displaystyle P(n)\) be the statement \(\displaystyle \displaystyle\sum_{i=0}^{n} i^3 =1^3 + 2^3 + ... + n^3 = (1+2+3+...+n)^2\)

The sum of the first \(\displaystyle n\) positive integers is \(\displaystyle \frac{n(n+1)}{2}\) which is a theorem refered to as 1-1.

We can rewrite the statement \(\displaystyle P(n)\) based on theorem 1-1 as \(\displaystyle \displaystyle\sum_{i=0}^{n} i^3 = (\frac{n(n+1)}{2})^2\).

Now we will show that \(\displaystyle P(n)\) is true for all integers \(\displaystyle n \geq 0\)

**Base case**\(\displaystyle P(0)\) is the statement \(\displaystyle \displaystyle\sum_{i=0}^{0} i^3 = (\frac{0(0+1)}{2})^2= 0\) which is clearly true.

**The induction step**Let \(\displaystyle k \geq 0\) be an integer. Assume (for induction) that \(\displaystyle P(k)\) is true. That means \(\displaystyle \displaystyle\sum_{i=0}^{k} i^3\) \(\displaystyle \stackrel{?}{=}\)\(\displaystyle (\frac{k(k+1)}{2})^2\), is true for \(\displaystyle n=k\).

We will prove that \(\displaystyle P(k+1)\) is true as well. That is, we must prove that \(\displaystyle \displaystyle\sum_{i=0}^{k} i^3 + (k+1)^3= (\frac{(k+1)((k+1)+1)}{2})^2\). To prove this equation, start by adding \(\displaystyle (k+1)^3\)

to both sides of the inductive hypothesis:

\(\displaystyle \displaystyle\sum_{i=0}^{k+1} i^3= (\frac{k(k+1)}{2})^2 + (k+1)^3\). Now, simplifying the right side we get:

\(\displaystyle =\frac{k^2((k+1)^2)}{4} + (k+1)^3 = \frac{k^2(k+1^2) +(4k+4)(k+1)^2}{4} = \frac{(k+2)^2(k+1)^2}{2^2} = (\frac{((k+2)(k+1)=}{2})^2 = (\frac{(k+1)((k+1)+1)}{2})^2 \)

Thus \(\displaystyle P(k+1)\) is true, so by the principle of mathematical induction \(\displaystyle P(n)\) is true for all integers \(\displaystyle n \geq 0\)

Q.E.D.

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