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

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

**Statement **
[tex]\displaystyle\sum_{i=

0}^{n} x^3 =1^3 + 2^3 + ... + n^3 = (1+2+3+...+n)^2[/tex]

By using theorem 1-1: (we don't know what Theorem 1-1 is)The sum of the first n positive integers is [tex]\frac{n(n+1)}{2}[/tex]

We can rewrite the statement as [tex]\displaystyle\sum_{i=

0}^{n}

x^3 =1^3 + 2^3 + ... + n^3 = (1+2+3+...+n)^2=(\frac{n(n+1)}{2})^2[/tex]

Show true for [tex]n=1[/tex]

[tex]\displaystyle\sum_{i=

0}^{1}

x^3 = 0^3 + 1^3 = 1[/tex] and [tex]1^2 = 1[/tex]

LHS(Left-hand side) = RHS(Right-hand side), therefore, the statement is true for [tex]n = 1[/tex]

**The induction hypothesis **
[tex]\displaystyle\sum_{i=

0}^{k+1} x^3[/tex] [tex]\stackrel{?}{=}[/tex] is true for some value of [tex]k≥n[/tex], hence [tex]k≥1[/tex].

(I would write true for n=k)

**The induction step **
We must show that the statement is true for some value of [tex]n = k +1[/tex], such that it will also be true for the next value of [tex]n[/tex].

(not very clear)
Assuming that [tex]\displaystyle\sum_{i=

0}^{k}

x^3=(\frac{k(k+1)}{2})^2[/tex], we find that

[tex]\displaystyle\sum_{i=0}^{k+1} x^3= \displaystyle\sum_{i=0}^{k} x^3 + (k+1)^3 = (\frac{k(k+1)}{2})^2 + (k+1)^3= [/tex]

[tex]\frac{k^2((k+1)^2)}{4} + (k+1)^

2 = \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) [/tex]

The last

equation is the same equation as [tex](\frac{n(n+1)}{2})^2[/tex] which is based on Theorem1-1, except that n is replaced by (k+1). Hence whenever it is true for all the integers 1,2,...,k, then it is true for the integer k+1.

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