# 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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In order for you to understand the linked questions, it is maybe worth mentioning that $\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. I hope you know that. – t.b. Sep 6 '11 at 2:45
@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

## 9 Answers

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 $B(n+1)$?

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

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@RajeshKSingh let me know if you have questions about the steps of the proof. – user2468 Aug 29 '12 at 23:08
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
@RajeshKSingh No. If you want to prove a statement about all natural numbers then you need induction. – user2468 Aug 29 '12 at 23:16
@JenniferDylan: the steps are clear to me. is there a method different from induction? – Rajesh K Singh Aug 29 '12 at 23:25
@JenniferDylan: thanks – Rajesh K Singh Aug 29 '12 at 23:26

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.

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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

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 $n = k+1$.

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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

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.)

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

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

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if you are interested we can discuss about the proof.! – Master X 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

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

This is the inductive step. The rest is easy.

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Let $S(n)$ be the statement: $1^{3}+2^{3}+\cdots+n^{3}=\bigg(\dfrac{n(n+1)}{2}\bigg)^{2}$; $n\in \mathbb{N}$

Basis step: $\hspace{5 mm}S(1)$:

LHS $=1^{3}+2{3}+\cdots+(1)^{3}$ $=1$

RHS $=\bigg((1)\dfrac{\big((1)+1\big)}{2}\bigg)^{2}$ $=1$

$\hspace{75 mm}$LHS $=$ RHS $\hspace{1 mm}$(verified.)

Inductive step:

Assume $S(k)$ is true, i.e. assume that $1^{3}+2^{3}+\cdots+k^{3}=\bigg(\dfrac{k(k+1)}{2}\bigg)^{2}$; $k\in \mathbb{N}$

$S(k+1)$: $1^{3}+2^{3}+\cdots+k^{3}+(k+1)^{3}=\bigg(\dfrac{k(k+1)}{2}\bigg)^{2}+(k+1)^{3}$

$\hspace{70 mm}=\dfrac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}$

$\hspace{70 mm}=\dfrac{k^{2}(k+1)^{2}+4(k+1)^{3}}{4}$

$\hspace{70 mm}=\dfrac{(k+1)^{2}(k^{2}+4k+4)}{4}$ $\hspace{70 mm}=\dfrac{(k+1)^{2}(k^{2}+2k+2k+4)}{4}$ $\hspace{70 mm}=\dfrac{(k+1)^{2}\big(k(k+2)+2(k+2)\big)}{4}$ $\hspace{70 mm}=\dfrac{(k+1)^{2}(k+2)(k+2)}{4}$ $\hspace{70 mm}=\dfrac{(k+1)^{2}(k+2)^{2}}{4}$

So, $S(k+1)$ is true whenever $S(k)$ is true.

Therefore, $1^{3}+2^{3}+\cdots+n^{3}=\bigg(\dfrac{n(n+1)}{2}\bigg)^{2}$; $n\in \mathbb{N}$.

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In the basis step, I used 1 because there is ambiguity surrounding the membership of 0 in the set of all natural numbers. – Tazwar Sikder Jun 13 at 14:59
In the verification step, you have "proven" that $1^3 + 2^3 + \dots + 0^3 = 0$. Does this look correct to you? Also, why answer a question that is 5 years old, that already has 8 answers (of which one accepted), that is quite easy, that is a community wiki (and thus brings you no reputation points), only to give a proof that is essentially identical to all the ones already existing? Given the downvotes that you have gained, if I were you I would delete this. – Alex M. Jun 14 at 15:51
Thanks for the tip. This is my first answer on Stack Exchange. – Tazwar Sikder Jun 14 at 18:18