# 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

• @Steve: See this answer for general comments on induction, and this one for specific advice on doing proofs by induction. The example there may be enough for you to figure out how to prove this statement by induction. – Arturo Magidin Sep 6 '11 at 3:34
• Since this question is asked quite frequently, it has been added to the list of Generalizations of Common questions. It has been kept seperate from the version which does not use induction. – Eric Naslund Aug 30 '12 at 0:23
• This answer has a great discussion of how to do an induction proof – Ross Millikan Oct 30 '17 at 21:09
• I would say this must have been asked on this website at least $10$ times. – Clement C. Apr 19 '18 at 21:14
• More similar questions with answers can be found with Approach0 – Martin R Oct 22 '19 at 11:49

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: If $(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$, then \begin{eqnarray} (1^3+2^3+3^3+\cdots+n^3 + (n+1)^3)&=&(1+2+3+\cdots+n)^2 +(n+1)^3\\ &=&(\dfrac{n^2(n+1)^2}{2} + (n+1)^3\\ &=& (n+1)^2(\dfrac{n^2}{2}+n+1)\\ &=& (n+1)^2((n+1)^2+1)/2\\ &=& (1+2+ \cdots +n +(n+1))^2 \end{eqnarray} – user29999 Aug 29 '12 at 23:14
• @user29999 You didn't square the denominator. D: – Simply Beautiful Art Dec 9 '16 at 21:44
• @user29999 You do not necessarily need induction, but its not a bad idea. – Simply Beautiful Art Dec 9 '16 at 21:45

Hint  First trivially inductively prove the Fundamental Theorem of Difference Calculus

$$\rm\ F(n) = \sum_{k\, =\, 1}^n f(k)\, \iff\, F(n) - F(n\!-\!1)\, =\, f(n),\ \ \, F(0) = 0\qquad$$

The result now follows immediately by $$\rm\ F(n) = (n\:(n\!+\!1)/2)^2\ \Rightarrow\ \color{#c00}{F(n)-F(n\!-\!1) = n^3}$$

The theorem reduces the proof to a trivial mechanical verification of a $$\rm\color{#c00}{polynomial\ equality}$$, which requires no ingenuity whatsoever.

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

• I got here from your latest post, how do you find $F(n)$ ? do you have a method or do you just 'see' this ? – Belgi Sep 19 '12 at 10:36
• @Belgi Here the sum closed form $\rm F(n)$ is given and we are merely asked to verify it correctness. Computing the closed form is known as summation in finite terms and there are known algorithms for wide-classes of special functions, e.g. chase citations to Michael Karr's papers on "Summation in finite terms". – Bill Dubuque Sep 4 '17 at 23:32

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 $n = k+1$.

• Is there any particular reason why you are writing in bold face? It's a bit distracting, and not the usual for this site... Note also that it doesn't work for math formulas, so that makes it even more distracting... – Arturo Magidin Sep 6 '11 at 4:14
• I wanted the math formulas to appear in bold face,but the opposite is happening.don't worry i'll edit it. – alok Sep 6 '11 at 4:17
• You marked not just the equations, also all the text. And why was it important for the formulas to be in boldface? Again, not the usual for this site, and they don't look the same: $k$ vs. $\mathbf{k}$, $+$ vs. $\mathbf{+}$, $\cdots$ vs. $\mathbf{\cdots}$. I've cleaned it up a bit. – Arturo Magidin Sep 6 '11 at 4:21

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

• This doesn't "show" it in the sense of a proof, though. Which is what the question asked for. – YiFan Oct 16 '19 at 23:48
• @YiFan:You are right, but it is another idea. That's proof without a word. – Khosrotash Oct 18 '19 at 3:47

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 $\frac{n^{r+1}}{r+1}$

• if you are interested we can discuss about the proof.! – user229886 Apr 9 '15 at 6:58
• you can also see page 84-85-,86 of proofs without word by roger B. Nelson ,ISBN 088385-700-6 – Khosrotash Jul 21 '15 at 6:51

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.)

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, $$(\sum_{i=1}^1 i)^2 = \sum_{i=1}^1 i^3=1$$

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

This is the inductive step. The rest is easy.

[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]

HINT

We need to prove

• Base case: set $i=1$ and check by inspection
• Induction step: assume $\sum_{i=0}^n i^3 = (\sum_{i=0}^n i)^2$ true prove

$$\sum_{i=0}^{n+1} i^3 = \left(\sum_{i=0}^{n+1} i\right)^2$$

Note that

• $\sum_{i=0}^{n+1} i^3=(n+1)^3+\sum_{i=0}^{n} i^3$
• $\left(\sum_{i=0}^{n+1} i\right)^2=\left(n+1+\sum_{i=0}^{n} i\right)^2$
• It helps to recall that $1+2+ \ldots +n = n(n+1)/2.$ – Chris Leary Apr 19 '18 at 21:20
• @ChrisLeary Oh yes, I supposed that should be well known, anyway that's good to recall! Thanks – user Apr 19 '18 at 21:23
• Sorry, that comment was for the OP, not you. I should have addressed it directly to Ryan. – Chris Leary Apr 20 '18 at 2:10

[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]

You should prove before $$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$

Cases $$n=0,1$$ are trivial. Suppose it's true for $$n-1$$ then \begin{align} \sum_{i=0}^n i^3 &= n^3 + \sum_{i=0}^{n-1}i^3 \\ &=n^3 + \left(\sum_{i=0}^{n-1} i\right)^2 \\ &= n^3 + \frac{n^2(n-1)^2}{4}\\ & = \frac{4n^3 + n^4 + n^2 + -2n^3}{4}\\ &= \frac{(n+1)^2n^2}{4} \\ &= \left(\sum_{i=0}^{n} i\right)^2 \end{align}

Hint: $$(1+2+...+n+(n+1))^2 - (1+2+...+n)^2 = 2(1+2+...+n)(n+1) + (n+1)^2 = (n+1)(2(1+2+...+n) + (n+1)) = (n+1)(n(n+1) + n+1) = (n+1)(n+1)^2 = (n+1)^3$$

We use here the (slightly easier to prove) result $$\sum_{k=1}^{n} k = n(n+1)/2$$

A historical way of computing: $$~$$ Recursion

$$(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$$

Add up for $$~k=1...n~$$ .

Now the formula for $$~\sum\limits_{k=1}^n k^3~$$ is based on $$~\sum\limits_{k=1}^n k^2~$$ , $$~\sum\limits_{k=1}^n k~$$ and $$~\sum\limits_{k=1}^n 1~$$ .

A direct way of computing:

$$\displaystyle\sum\limits_{k=1}^n k^3 = \sum\limits_{j=1}^n\sum\limits_{k=j}^n k^2 =\sum\limits_{j=1}^n\sum\limits_{k=j}^n jk + \sum\limits_{j=1}^n\sum\limits_{k=j}^n (k-j)k$$

$$\displaystyle\hspace{1.3cm}=\sum\limits_{j=1}^n\sum\limits_{k=j}^n jk + \sum\limits_{j=1}^n\sum\limits_{k=1}^{j-1} kj =\sum\limits_{j=1}^n j\sum\limits_{k=1}^n k = \left(\sum\limits_{j=1}^n k\right)^2$$

Transformation rule: $$~(k-j)k|_{j=a\\k=a+b}=kj|_{j=a+b\\k=b}~$$ with $$~a,b,a+b\in\{1,2,...,n\}$$

Other ways of computing: Faulhaber's formula, $$~\sum\limits_{k=1}^n k^m x^k = \left(x\frac{d}{x}\right)^m \left(\frac{x^{n+1}-1}{x-1}\right)~$$ , Euler-Maclaurin_formula , Interpolation ($$5$$ interpolation points to get a polynomial of degree $$4$$, here for $$~\sum\limits_{k=1}^n k^3 ~$$), integration-like calculation (see Discrete Mathematics), and so on ...

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

\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}

• It seems that induction has been asked for. – Claude Leibovici Jul 9 '16 at 2:57
• The question clearly asks for a proof using P.M.I. – Shubh Khandelwal Aug 24 '18 at 4:18

We can try by induction, that is

• base case: $$1^3=1^2$$
• induction step: assume $$1^3 + 2^3 +...+ n^3 = (1 + 2 +...+ n)^2=\frac{n^2(n+1)^2}{4}$$ then

$$1^3 + 2^3 +...+ n^3+(n+1)^3=\frac{n^2(n+1)^2}{4}+(n+1)^3=\\=(n+1)^2\left(\frac{n^2}{4}+(n+1)\right)=(n+1)^2\frac{n^2+4n+4}{4}=\frac{(n+1)^2(n+2)^2}{4}$$

• This has been asked and answered before many times. A possible duplicate target has already been pointed out. – Martin R Oct 22 '19 at 12:07

They are both polynomials of degree $$4$$. To prove that they are equal, it is enough to prove that their values coincide for $$5$$ values of $$n$$, for instance $$n=0,1,2,3,4$$.
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