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so we're looking to prove $P(n)$ that

$$1^2+2^3+\cdots+n^3 = (n(n+1)/2)^2$$

I know the basis step for $p(1)$ holds.

We're going to assume $P(k)$

$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$

And we're looking to prove $P(k+1)$

What I've discerned from the internet is that I should be looking to add the next term, $k+1$, to both sides so...

$$1^3+2^3+\cdots+k^3 + (k+1)^3=(k(k+1)/2)^2 + (k+1)^3$$

now I saw some nonsense since that we assumed $p(k)$ we can use it as a definition in our proof, specifically on the left hand side

so since $$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$

then $$(k(k+1)/2)^2 + (k+1)^3 = (k(k+1)/2)^2 + (k+1)^3$$ and we have our proof

OK so far thats wrong

so far ive figured this.

$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+1)+1)/2)^2$$

Then

$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+2)/2)^2$$

using the definition

$$(k(k+1)/2)^2 + (k+1)^3 = ((k+1)((k+2)/2)^2$$ $$(k^2+k/2)^2 + (k^2+2k+1)(k+1) = (k^2+3k+2/2)^2$$ $$(k^4+k^2/4)+(k^2+2k^2+k+k^2+2k+1)= (k^4+9k^2+4/4)$$

Where should I go from here? It doesn't possibly look like these could equate, I'll keep going though

Michael Hardy
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Mike
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3 Answers3

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Assuming $P(k)$, you add $(k+1)^3$ on both sides of $$ 1^3 + 2^3 + \ldots + k^3 = (k(k+1)/2)^2 $$ to get \begin{align} 1^3 + 2^3 + \ldots + k^3 + (k+1)^3 & = (k(k+1)/2)^2 + (k+1)^3 \\ & = \frac 14 k^2(k+1)^2 + (k+1)^3 \\ & = \frac 14\left(k^4 + 2k^3 + k^2 + 4k^3 + 12k^2 + 12k + 4\right) \\ & = \frac 14\left(k^4 + 6k^3 + 13k^2 + 12k + 4\right) \\ & = \frac 14(k+1)^2(k+2)^2 \\ & = \left((k+1)(k+2)/2\right)^2. \end{align} This statement is $P(k+1)$.

Tunococ
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What you need to show is that $S(k-1)+k^3=S(k)$, i.e. $$\frac{(k-1)^2k^2}4+k^3=\frac{k^2(k+1)^2}4.$$ Simplifying by $\frac{k^2}4$, you get $$(k-1)^2+4k=(k+1)^2.$$ QED.

  • This is an abuse of notation. $P(k)$ was used to denote the statement that $\sum\limits_{i = 1}^k i^3 = \left(\frac{k(k + 1)}{2}\right)^2$. Adding $(k + 1)$ to a proposition doesn't make sense. Better would be to say that the goal is to show that for all $k \in \mathbb N$, $P(k) \implies P(k + 1)$, which can be done by assuming $P(k)$, adding $(k + 1)^3$ to both sides of the equality given by $P(k)$, and simplifying to give that $P(k + 1)$ holds. – qaphla Jun 30 '14 at 22:52
  • How is that a finished proof? and i guess I kind of understand how you got the $$(k+1)^2$$ out of there. How are these equivalent? – Mike Jun 30 '14 at 23:22
  • I pretty much have that going to $$k^2/4 + 4(k+1)/4 = k^2+4/4$$ and then $$k^2+4k+4/4=k^2+4/4$$ – Mike Jun 30 '14 at 23:25
  • Sorry, I meant $S(k)$, the sum of the first $k$ cubes (not the proposition $P(k)$). –  Jun 30 '14 at 23:34
  • Can anyone help me out with this last part? – Mike Jun 30 '14 at 23:34
  • Hum, $(\frac{k+2}2)^2$. –  Jun 30 '14 at 23:36
  • unfortunately that doesn't make a whole lot of sense. I guess my main confusion is figuring out what my goal is to prove? Like I know I'm looking for s(k) + (k+1)^3=s(k+1) but in the end they don't equate? I'm missing something, bear with me here – Mike Jun 30 '14 at 23:55
  • They are equal. Double check your derivation. I gave you the hint. –  Jul 01 '14 at 01:26
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An example proof:

Let $P(k)$ denote the statement that $\sum\limits_{i = 1}^k i^3 = \left( \frac{k(k + 1)}{2} \right)^2$.

We wish to prove that for all $k \in \mathbb Z$ such that $k > 0$, $P(k)$.

We will prove this by induction on $k$.

In the base case, $k = 1$, and we have that $1^3 = 1 = \left(\frac{1(1 + 1)}{2}\right)^2$.

Now, for the inductive case, we suppose that for some $k$, $P(k)$ holds.

Then, as we have assumed $P(k)$, we have that $\sum\limits_{i = 1}^k i^3 = \left(\frac{k(k + 1)}{2} \right)^2$.

Adding $(k + 1)^3$ to both sides, we get that $\sum\limits_{i = 1}^{k + 1} i^3 = \left(\frac{k(k + 1)}{2} \right)^2 + (k + 1)^3$.

Now, the right-hand side is equal to, by factoring, $(k + 1)^2 \left(\frac{k^2}{4} + (k + 1)\right)$.

We can then rewrite this as $(k + 1)^2 \left(\frac{k^2 + (4(k + 1))}{4}\right)$, which then becomes $\frac{(k + 1)^2(k^2 + 4k + 4)}{4}$, or $\frac{(k + 1)^2(k + 2)^2}{4}$.

This is equal to $\left(\frac{(k + 1)(k + 2)}{2}\right)^2$, and so we have that $$ \sum\limits_{i = 1}^{k + 1} i^3 = \left(\frac{(k + 1)(k + 2)}{2}\right)^2, $$ which is precisely $P(k + 1)$.

Thus, we have shown that $P(1)$ holds and that for all $k \in \mathbb Z$ such that $k > 0$, $P(k) \implies P(k + 1)$, and so, by the principle of mathematical induction, $P(k)$ holds for all $k$.

qaphla
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