# 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

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

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

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

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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