Help with equation manipulation when performing inductive proof.

Question

Can someone please assist me with what I feel is a foolishly simple concept. I have returned to school after many years, and my algebra skills seem to have deteriorated significantly. We are studying inductive proofs, and while I actually grasp the general concept of the proof, I am struggling with the algebra required.

We have been given the following example in our homework:

I am having difficulty following the progression of these steps. I am fairly certain the instructor has skipped a couple steps and not show what is to him "intuitive". Can someone please refresh my skills on how this works out including the factoring?

Answer 1

I suppose that this is what you search :

$$ \left(\frac{m(m+1)}{2} \right)^2+(m+1)^3= (m+1)^2\left(\frac{m}{2}\right)^2+(m+1)^3= $$ since $(m+1)^2$ is a ''common factor'' we have: $$ =(m+1)^2\left(\frac{m^2}{4}+(m+1) \right)=\frac{(m+1)^2}{4}\left(m^2+4(m+1) \right)= $$ $$ =\left(\frac{m+1}{2} \right)^2\left(m^2+4(m+1) \right) $$

Answer 2

From first line to second line: this is directly the inductive hypothesis.

Second to third line: factor out $\left(\frac{m+1}{2}\right)^2$ to obtain $$\left(\frac{m+1}{2}\right)^2\left[m^2+(m+1)^3 \times \left(\frac{m+1}{2}\right)^{-2}\right]$$ Then note that $(m+1)^3 \times (m+1)^{-2} \times \left(\frac{1}{2}\right)^{-2} = (m+1) \times 4$.

The trip from third line to fourth line writes $m^2+4(m+1)$ as $(m+2)^2$ and then brings it inside the bracket of $\left(\frac{m+1}{2}\right)^2$.

Answer 3

You should prove that $\sum_{k=0}^mk^3 = \frac{(m)^2(m+1)^2}{4}$ using induction.

first you should prove it is true for the base case ($k=0$),

ie $\sum_{k=0}^0k^3 =\frac{(0)^2(0+1)^2}4 = 0 $

Hence it is true .

Next step is the inductive step where you prove the ($m+1$)$^{th}$ term using the $m^{th}$ term.

$\sum_{k=0}^{m+1}k^3 =\sum_{k=0}^mk^3+(m+1)^3$

$=\frac{(m^2)(m+1)^2}{4}+(m+1)^3 $

(take $(m+1)^2 $ as common from the terms.)

$=(m+1)^2.\bigg(\frac{m^2}4+m+1\bigg) $

$= \frac{(m+1)^2}{4}.\bigg(m^2+4(m+1)\bigg)$

$=\frac{(m+1)^2(m+2)^2}{4}$

$= \frac{(m+1)((m+1)+1)}{4}$

hence $m+1\implies m$

the inductive proof is then considered complete.