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Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction

My proof so far:

Let $P(n)$ be $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$


Base Case

$P(1):$

LHS = $1^3 = 1$

RHS = $(1)^2 = 1$

Since LHS = RHS, therefore base case holds


Induction Hypothesis

Let $n \in \mathbb{N}$ be arbitrary

Assume $P(n)$ holds


Induction Step

Prove $P(n+1)$ holds:

$$ \begin{align} & 1^3 + 2^3 + \cdots + \;n^3 + (n+1)^3 \\ = {} & (1 + 2 + \cdots + \; n)^2 + (n+1)^3 \text{ (by Induction Hypothesis)} \\ = {} & (1 + 2 + \cdots + \; n)^2 + (n^3 + 3n^2 + 3n + 1) \end{align} $$


This is where I get stuck. I don't know how to prove that my last step is equivalent to:

$$(1 + 2 + \cdots + \;n + (n+1))^2$$

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marked as duplicate by Adam Hughes, Jonas Meyer, Thomas Andrews, dustin, user147263 Jan 20 '15 at 3:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Thank you for linking it. I did not notice that. I can now see additional takes at the question. $\endgroup$ – ashimashi Jan 20 '15 at 3:52
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So basically you want that last expression to turn out to be $\big(1+2+\ldots+n+(n+1)\big)^2$, so you want $(n+1)^3$ to be equal to the difference

$$\big(1+2+\ldots+n+(n+1)\big)^2-(1+2+\ldots+n)^2\;.$$

That’s a difference of two squares, so you can factor it as

$$(n+1)\Big(2(1+2+\ldots+n)+(n+1)\Big)\;.\tag{1}$$

To show that $(1)$ is just a fancy way of writing $(n+1)^3$, you need to show that

$$2(1+2+\ldots+n)+(n+1)=(n+1)^2\;.$$

Do you know a simpler expression for $1+2+\ldots+n$?

Work forward from this...

Okay so going forward.

Basically, Induction works like this, I'll use your question as an example:

Consider the case when $n = 1$. If so, then we will have $1^3 = 1^2$.

Now suppose that $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$.

Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$.

Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$.

Hence, the statement holds for the $n + 1$ case. Thus by the mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.

$\mathbb Q.\mathbb E.\mathbb D$

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  • $\begingroup$ @ashimashi No Problem! $\endgroup$ – Yagna Patel Jan 20 '15 at 3:52
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Hint: by the binomial theorem,

$(1+2+\cdots +n +(n+1))^2=(1+2+\cdots +n)^2+(n+1)^2+2(n+1)(1+2+ \cdots n)$.

Now, you can show by induction (or by Gauss' trick) that $1+2+\cdots +n= \dfrac{n(n+1)}{2}$, and hus reach the answer.

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    $\begingroup$ For some reason, it skipped my mind that I could have written it as $\frac{n(n+1)}{2}$. Thanks a lot! $\endgroup$ – ashimashi Jan 20 '15 at 3:48

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