Prove by induction $\forall n\in N, \sum_{k=1}^nk^3=(\sum_{k=1}^nk)^2$

Question

Step 1 of Induction: Prove base case
$1^3=1^2$

Since we've proven base case we can assume
$\forall n\in N, \sum_{k=1}^nk^3=(\sum_{k=1}^nk)^2$
Is true

Now need to prove
$$\sum_{k=1}^{n+1}k^3=(\sum_{k=1}^{n+1}k)^2$$ $$\sum_{k=1}^{n+1}k^3=\sum_{k=1}^nk^3+(n+1)^3$$ This is where I get stuck. $(\sum_{k=1}^{n+1}k)^2$ is kind of tricky to get into an equivalent form. I can see that $(\sum_{k=1}^{n+1}k)^2=(\sum_{k=1}^{n}k+(n+1))^2$, but I don't know how to proceed from here.

Try expanding $(\sum_{k=1}^{n}k+(n+1))^2$ using the binomial theorem — J. W. Tanner, Oct 12 '19 at 20:46
Hint - write $(\sum_{k=1}^{n+1}k)^2=\left(\dfrac{(n+1)(n+2)}{2}\right)^2$. — M.P, Oct 12 '19 at 20:47
that's kind of lame the question doesn't mention $\sum k^3=[n(n+1)/2]^2$ — Gooby, Oct 12 '19 at 20:49
@DylanY so you don't know that $1+2+...+n=\frac{n(n+1)}{2}$? — rtybase, Oct 12 '19 at 21:02
It's something if I see I remember is true, but don't have memorized by heart. — Gooby, Oct 12 '19 at 21:03
Well, it's a well known topic and as such it was asked before, thus it's a duplicate ... — rtybase, Oct 12 '19 at 21:07

score 1 · Accepted Answer · answered Oct 12 '19 at 20:53

Now $$(\sum_{k=1}^{n+1}k)^2= [(\sum_{k=1}^{n}k)+(n+1)]^2=(\sum_{k=1}^{n}k)^2+2(n+1)\sum_{k=1}^{n}k+(n+1)^2=\sum_{k=1}^{n}k^3+2(n+1)\sum_{k=1}^{n}k+(n+1)^2=...=\sum_{k=1}^{n}k^3+(n+1)^3$$

So what we want to prove is that $$(n+1)^3=2(n+1)\sum_{k=1}^{n}k+(n+1)^2\\(n+1)^3-(n+1)^2=2(n+1)\sum_{k=1}^{n}k \\ (n+1)^2(n+1-1)=2(n+1)\sum_{k=1}^{n}k\\ \frac{n(n+1)}{2}=\sum_{k=1}^nk$$

and the last row is the well-known formula of the sum of the first $n$ natural numbers

Prove by induction $\forall n\in N, \sum_{k=1}^nk^3=(\sum_{k=1}^nk)^2$

1 Answers1