Inductive Method

Question

I need to test by inductive method $1^3 + 2^3 + 3^3 + \dots + n^3 = (1 + 2 + 3 + \dots +n)^2.$

1) base equation $$\sum_{i=1}^{n}{i^{3}} = { (n + 1) ^{2}}$$ 2) replace with 1 $$\sum_{i=1}^{n}{i^{3}} = { (1+1) ^{2}}$$ Is that correct?

3) Assumption: $$\sum_{i=1}^{k}{i^{3}} = { (k+1) ^{2}}$$ 4) Check: \begin{align*} \sum_{i=1}^{k+1}{i^{3}} &= { ((k+1) +1) ^{2}}\\ \sum_{i=1}^{k}{i^{3}} + (k+1)&= ((k+1)+1) ^{2} (k+1) ^{2} + (k+1) \\ &= ((k+1)+1) ^{2} (k+1) ^{2} + (k+1)\\&= { ((k+1)^{2}+2k+2+1)}\end{align*} and can not continue more....

Answer 1

This statement is false. See this graph, the only intersection is when $x=2$, for the rest, it doesn't work.

E.g. for $x=3$, $\sum_{i=1}^{3}{i^3}=1^3+2^3+3^3=1+8+27=36$, while $(3+1)^2=16$.

What your base statement should actually be is: $$\sum_{i=1}^{n}{i^3}=\bigg[\sum_{i=1}^{n}{i}\bigg]^2$$

For the result: prove these by induction: $$\sum_{i=1}^{n}{i}=\frac{1}{2}n(n+1)$$ $$\sum_{i=1}^{n}{i^3}=\frac{1}{4}n^2(n+1)^2$$

If you can manage that, it's easy to see that $$\sum_{i=1}^{n}{i^3}=\bigg[\sum_{i=1}^{n}{i}\bigg]^2$$

Answer 2

It is important to note when proving by induction to verify the left and right hand sides independently of one another, otherwise you can mistakenly prove an incorrect statement by assuming equality.

The proof for this statement would go as follows:

Base Step: n=1

$\sum_{i=1}^{n=1}i^3=1^3=1$ and since $(1)^{2}=1$, we may proceed to the inductive step.

Inductive Step:

Assume $\sum_{i=1}^{k}i^3=(1+2+\cdots+k)^2$ for some $\it{k\in \mathbb{N}}$. Then for ${k+1}$, we have

$\sum_{i=1}^{k+1}i^3=\sum_{i=1}^{k}i^3 +(k+1)^3=(1+2+\cdots+k)^2+(k+1)^3=\left(\frac{(k+1)(k)}{2}\right)^{2}+(k+1)^3$

$=\left(\frac{k^{2}+k}{2}\right)^2+k^3+3k^2+3k+1=\left(\frac{k^{4}+3k^{3}+k^2}{4}\right)+\left(\frac{4k^{3}+12k^2+12k+4}{4}\right)=\left(\frac{k^4+6k^3+13k^2+12k+4}{2}\right)$

We return to our right side,

$(1+2+\cdots+k+k+1)^2=\left(\frac{(k+2)(k+1)}{2}\right)^2=\left(\frac{k^2+3k+2}{2}\right)^2=\left(\frac{k^4+6k^3+13k^2+12k+4}{4}\right)$

and we see we've obtained equality. Therefore $\sum_{i=1}^{n}i^3=(1+2+\cdots+n)^2\quad \forall n\in \mathbb{N}. \Box$

To understand why $(1+2+\cdots+k)=\left(\frac{(k+1)(k)}{2}\right)=\binom{k+1}{2}$ requires another proof by induction, but you can also search the internet for 'triangle number' for a brief overview.