# Proving by induction that $1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$

Need guidance on this proof by mathematical induction. I am new to this type of math and don't know how exactly to get started.

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

• What's PMI? What does "n3" mean? ... – DonAntonio Mar 25 '13 at 23:31
• I'm assuming it's "proof by mathematical induction" and the question is asking about the result for the sum of the first $n$ cubes. – Stahl Mar 25 '13 at 23:32
• Show that the formula is true for $n=1$. assume the formula is true for $n = k$. using that information, show that the formula holds true for $k + 1$. These sorts of proofs by induction are very straightforward, but only if you know how an induction proof works. – Tyler Mar 25 '13 at 23:35
• Possible duplicate – Adi Dani Mar 26 '13 at 0:03
• @AdiDani huh? what does that mean – JustaBreitGuy Mar 26 '13 at 0:08

To prove something via mathematical induction, first you need to identify the mathematical statement $P(n)$ that you're trying to verify. By mathematical statement, I mean something like "let $P(n)$ be the statement $n = n$." When you prove something with induction, you are trying to show that $P(n)$ is true for all $n\in\Bbb{N}$. See if you can identify the statement relevant to your problem first, but if you get stuck, you can hover your mouse over the area below to see it.

Let $P(n)$ be the statement: $\sum_{k = 1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$

The next step is showing that $P(1)$ is true; i.e. that the statement holds for $n = 1$. This should be easy.

$P(1):$ $\sum_{k = 1}^1 k^3 = 1^3 = \left(\frac{1\cdot2}{2}\right)^2 = 1^2 = 1$ is true.

After that, assume $P(n)$ is true. Then you want to use that assumption to show that $P(n+1)$ is true (you should be able to break the statement up into what you already know ($P(n)$) and then perform algebraic manipulation to show that $P(n+1)$ is true). Once you do that, the principle of mathematical induction tells you that $P(n)$ is true for all $n$, as $P(1)$ is true and $P(n)\implies P(n+1)$. I've left the last part for you to figure out yourself, as it's the bulk of the proof.

Hint: Start with $P(n)\implies \sum_{k = 1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$, so $\sum_{k = 1}^{n+1} k^3 = \left[\frac{n(n+1)}{2}\right]^2 + \left(n+1\right)^3$.

This is easy for $n = 1$. Assume it is true for $n$. Then $$\sum_{k=1}^{n+1} k^3 = (n + 1)^3 + \sum_{k=1}^n k^3.$$ Now invoke the induction hypothesis to get $$\sum_{k=1}^{n+1} k^3 = (n + 1)^3 + {n^2(n + 1)^2\over 4}.$$ Now wrangle the right hand side until you get $${(n+1)^2(n+2)^2\over 4}.$$

Leaving out the trivial case, assume true for $n$, let's prove for $n+1$:

\begin{align*} 1^3+ \dots + n^3+(n+1)^3 &= \left(\frac{n(n+1)}{2}\right)^2+(n+1)^3 \\ & = (n+1)^2\left(\frac{n^2}{4}+n+1\right) \\ & = \left(\frac{(n+1)}{2}\right)^2(n^2+4n+4)\\ & = \left(\frac{(n+1)(n+2)}{2}\right)^2 \end{align*}

• @leo what do I do about the n+2? isnt it supposed to equal [n(n+1)/2]^2 – JustaBreitGuy Mar 26 '13 at 0:41
• No, the proof is for the (n+1) case so it is one more than the previous case. – JB King Mar 26 '13 at 0:46
• @JB King $$1^3+2^3+\ldots+n^3+(n+1)^3+(n+2)^3=\left[\frac{(n+2)(n+3)}{2}\right]^2$$ – leo Mar 26 '13 at 1:37

Inductive proofs of sums like yours are easily tackled using a very simple general method known as telescopy - see the trivial inductive proof here, of the following fundamental

Theorem $\rm \displaystyle\ \ \sum_{i\,=\,0}^n\, f(i)\, =\, g(n)\iff f(0) = g(0)\ {\rm\ and\ }\ f(n) \,=\, g(n)-g(n\!-\!1)\:\$ for $\rm\,n \ge 1.$

In your case we have $\rm\ f(n) \,=\, n^3,\,\ \ g(n) \,=\, (n(n+1)/2)^2\$ so $\rm\ f(0) = g(0)\$ and

$$\rm g(n)\!-\!g(n\!-\!1)\, =\, \dfrac{(n(n\!+\!1))^2}4 - \dfrac{((n\!-\!1)n)^2}4 =\, \dfrac{n^2\,\big((n\!+\!1)^2\!-(n\!-\!1)^2\big)}{4} =\, n^3\!= f(n)\ \ \ {\bf QED}$$

You can find many more examples of telescopy and related results in other answers here.