Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction

How can I prove that

$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$

for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.

Thanks

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You are trying to prove something of the form, $$A=B.$$ Well, both $A$ and $B$ depend on $n$, so I should write, $$A(n)=B(n).$$ First step is to verify that $$A(1)=B(1).$$ Can you do that? OK, then you want to deduce $$A(n+1)=B(n+1)$$ from $A(n)=B(n)$, so write out $A(n+1)=B(n+1)$. Now you’re trying to get there from $A(n)=B(n)$, so what do you have to do to $A(n)$ to turn it into $A(n+1)$, that is (in this case) what do you have to add to $A(n)$ to get $A(n+1)$? OK, well, you can add anything you like to one side of an equation, so long as you add the same thing to the other side of the equation. So now on the right side of the equation, you have $B(n)+{\rm something}$, and what you want to have on the right side is $B(n+1)$. Can you show that $B(n)+{\rm something}$ is $B(n+1)$?

Let the induction hypothesis be
$$(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$
Now consider:
$$(1+2+3+\cdots+n + (n+1))^2$$
\begin{align} & = \color{red}{(1+2+3+\cdots+n)^2} + (n+1)^2 + 2(n+1)\color{blue}{(1+2+3+\cdots+n)}\\ & = \color{red}{(1^3+2^3+3^3+\cdots+n^3)} + (n+1)^2 + 2(n+1)\color{blue}{(n(n+1)/2)}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^2 + n(n+1)^2}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^3} \end {align}
QED

HINT $\:$ First trivially inductively prove the Fundamental Theorem of Difference Calculus

$$\rm\ F(n)\ =\ \sum_{i\: =\: 1}^n\:\ f(i)\ \ \iff\ \ \ F(n) – F(n-1)\ =\ f(n),\quad\ F(0) = 0$$

The result now follows immediately by $\rm\ F(n)\ =\ (n\:(n+1)/2)^2\ \Rightarrow\ F(n)-F(n-1)\ =\: n^3\:.\$

Note that by employing the Fundamental Theorem we have reduced the proof to the trivial mechanical verification of a polynomial equation. No ingenuity is required.

Note that the proof of the Fundamental Theorem is much more obvious than that for your special case because the telescopic cancellation is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. For further discussion see my many posts on telescopy.

Let P(n) be the given statement. You’ll see why in the following step.
$$P(n):1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$

Step 1. Let $n = 1$.

Then $\mathrm{LHS} = 1^3 = 1$, $\mathrm{RHS} = \frac{1^2(1+1)^2}{4} = \frac{4}{4} = 1$.

So LHS = RHS, and this means P(1) is true!

Step 2. Let $P(n)$ be true for $n = k$; that is,
$$1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4}$$

We shall show that $P(k+1)$ is true too!

Add $(k+1)^3$, which is $(k+1)^{\mathrm th}$ term of the LHS to both sides of (1); then we get:
\begin{align*} 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\ &= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\ &= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\ &= \frac{(k+1)^2(k+2)^2}{4}.\\ 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{(k+1)^2(k+2)^2}{4} \end{align*}
I think this statement is the same as $P(n)$ with $n = k+1$.

For proof by induction; these are the $\color{green}{\mathrm{three}}$ steps to carry out:

Step 1: Basis Case: For $i=1 \implies \sum^{i=k}_{i=1} i^3=\frac{1^2 (1+1)^2}{4}=\cfrac{2^2}{4}=1$. So statement holds for $i=1$.

Step 2: Inductive Assumption: Assume statement is true for $i=k$: $$\sum^{i=k}_{i=1} i^3=\frac{k^2 (k+1)^2}{4}$$

Step 3: Prove Statement holds for $i=k+1$. You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^3=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$
To do this you cannot use: $$\sum^{i=n}_{i=1} i^3=\color{red}{\frac{n^2 (n+1)^2}{4}}$$ as this is what you are trying to prove.

So what you do instead is notice that:
$$\sum^{i=k+1}_{i=1} i^3= \underbrace{\frac{k^2 (k+1)^2}{4}}_{\text{sum of k terms}} + \underbrace{(k+1)^3}_{\text{(k+1)th term}}$$
$$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{1}{4}k^2+(k+1)\right)$$
$$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{k^2+4k+4}{4}\right)$$
$$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{(k+2)^2}{4}\right)=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$

Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.

(QED is an abbreviation of the Latin words “Quod Erat Demonstrandum” which loosely translated means “that which was to be demonstrated”. It is usually placed at the end of a mathematical proof to indicate that the proof is complete.)

If $S_r= 1^r+2^r+…+n^r=f(n)$ and

$\sigma =n(n+1)$ & $\sigma’=2n+1$

$S_1=\frac{1}{2}\sigma$

$S_2=\frac{1}{6}\sigma\sigma’$

$S_3=\frac{1}{4}\sigma^2$

$S_4=\frac{1}{30}\sigma\sigma'(3\sigma-1)$

$S_5=\frac{1}{12}\sigma^2(2\sigma-1)$

$S_6=\frac{1}{42}\sigma\sigma'(3\sigma^2-3\sigma+1)$

$S_7=\frac{1}{24}\sigma^2(3\sigma^2-4\sigma+2)$

$S_8=\frac{1}{90}\sigma\sigma'(5\sigma^3-10\sigma^2+9\sigma-3)$

$S_9=\frac{1}{20}\sigma^2(2\sigma^3-5\sigma^2+6\sigma-3)$

$S_{10}=\frac{1}{66}\sigma\sigma'(3\sigma^4-10\sigma^3+17\sigma^2-15\sigma+5)$

proof of this is based on the theorem
if r is a positive integer, $s_r$ can be expressed as a polynomial in $n$ of which the highest term in $\frac{n^{r+1}}{r+1}$

It may be helpful to recognize that both the RHS and LHS represent the sum of the entries in a the multiplication tables. The LHS represents the summing of Ls (I’ll outline those shortly), and the RHS, the summing of the sum of the rows [or columns])$$\begin{array}{lll} \color{blue}\times&\color{blue}1&\color{blue}2\\ \color{blue}1&\color{green}1&\color{red}2\\ \color{blue}2&\color{red}2&\color{red}4\\ \end{array}$$
Lets begin by building our multiplication tables with a single entry, $1\times1=1=1^2=1^3$. Next, we add the $2$s, which is represented by the red L [$2+4+2 = 2(1+2+1)=2\cdot2^2=2^3$].
So the LHS (green 1 + red L) currently is $1^3+2^3$, and the RHS is $(1+2)+(2+4)=(1+2)+2(1+2)=(1+2)(1+2)=(1+2)^2$.
$$\begin{array}{llll} \color{blue}\times&\color{blue}1&\color{blue}2&\color{blue}3\\ \color{blue}1&\color{green}1&\color{red}2&\color{maroon}3\\ \color{blue}2&\color{red}2&\color{red}4&\color{maroon}6\\ \color{blue}3&\color{maroon}3&\color{maroon}6&\color{maroon}9\\ \end{array}$$
Next, lets add the $3$s L. $3+6+9+6+3=3(1+2+3+2+1)=3\cdot3^2=3^3$. So now the LHS (green 1 + red L + maroon L) currently is $1^3+2^3+3^3$, and the RHS is $(1+2+3)+(2+4+6)+(3+6+9)=(1+2+3)+2(1+2+3)+3(1+2+3)=(1+2+3)(1+2+3)=(1+2+3)^2$.

By now, we should see a pattern emerging that will give us direction in proving the title statement.

Next we need to prove inductively that $\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and use that relationship to show that $1+2+3+\dots+n+\dots+3+2+1 = \dfrac{n(n+1)}{2}+ \dfrac{(n-1)n}{2} = \dfrac{n((n+1)+(n-1))}{2}=\dfrac{2n^2}{2}=n^2$

Finally, it should be straight forward to show that:
$$\begin{array}{lll} (\sum^n_{i=1}i+(n+1))^2 &=& (\sum^n_{i=1}i)^2 + 2\cdot(\sum^n_{i=1}i)(n+1)+(n+1)^2\\ &=& \sum^n_{i=1}i^3 + (n+1)(\sum^n_{i=1}i + (n+1) + \sum^n_{i=1}i)\\ &=& \sum^n_{i=1}i^3 + (n+1)(n+1)^2\\ &=& \sum^n_{i=1}i^3 + (n+1)^3\\ &=& \sum^{n+1}_{i=1}i^3\\ \end{array}$$
and, as was already pointed out previously, $$(\sum_{i=1}^1 i)^2 = \sum_{i=1}^1 i^3=1$$

$$\ S =\sum_{k=1}^n k^3$$

\begin{align}
k^3 =& Ak(k+1)(k+2)+Bk(k+1)+Ck+D \\
k^3= & Ak^3+(3A+B)k^2+(2A+B+C)k+D
\end{align}
Therefore $A=1$, $B=-3$, $C=1$, $D=0$ and

\begin{align}
S =& \sum_{k=1}^n (k(k+1)(k+2)-3k(k+1)+k)\\
S =&\sum_{k=1}^n k(k+1)(k+2)-3\sum_{k=1}^nk(k+1)+\sum_{k=1}^nk \\
S =&\sum_{k=1}^n6\binom{k+2}{3}-3\sum_{k=1}^n2\binom{k+1}{2}+\sum_{k=1}^n\binom{k}{1}\\
S=&6\binom{n+3}{4}-6\binom{n+2}{3}+\binom{n+1}{2}\\
S=&\left\lgroup\frac{n(n+1)}{2}\right\rgroup^2.
\end{align}

This picture shows that $$1^2=1^3\\(1+2)^2=1^3+2^3\\(1+2+3)^2=1^3+2^3+3^3\\(1+2+3+4)^2=1^3+2^3+3^3+4^3\\$$ this is handmade of mine

$$n^3=S_n-S_{n-1}=\left(\frac{n(n+1)}2\right)^2-\left(\frac{n(n-1)}2\right)^2=n^2\left(\frac{n+1}2-\frac{n-1}2\right)\left(\frac{n+1}2+\frac{n-1}2\right)\\ =n^2\cdot1\cdot n.$$

This is the inductive step. The rest is easy.