# Mathematical Induction Algebra simplification? (1 Viewer)

#### cchan334

##### Member
So usually for mathematical induction after you assume true for any integer k, and then you try and prove for k+1, and sub in the RHS, it's some easy algebra manipulation and they end up being equal. I don't know why but I've been staring at this question for 20 minutes and I can't seem to be able to do that. It seems that the only way to prove that it is true is by expanding both sides of the equation???

#### xilbur

##### New Member
(2(k+1)-1)^3 = (2k +2 - 1)^3
= (2k +1 )^3

Is that where you were stuck at?

If you're going straight from the assumption and doing LHS = RHS, be wary of expanding that k+1!

#### cchan334

##### Member
(2(k+1)-1)^3 = (2k +2 - 1)^3
= (2k +1 )^3

Is that where you were stuck at?

If you're going straight from the assumption and doing LHS = RHS, be wary of expanding that k+1!
I expanded it already, but like isn't there an easier way than brute expanding it because usually idk with some algebra manipulation I can show that they are equal without expanding it???

#### cossine

##### Member
You are almost there.

You wrote (2(k+1) -1)^3 as (2k-1)^3 instead of (2(k+1) -1)^3 as (2k+1)^3 as mentioned by xilbur.

From there it should be an easy direct proof.

Note: Make sure to prove true for n=1 and then assume true for n=k. Leaving these steps out might lead to deduction of marks

#### cchan334

##### Member
Oh my bad. I did the proof by brute expansion on both sides but I felt like there had to be an easier way?
Like is the only method for showing the LHS =RHS through expanding both sides cas I really can't see it being done any other way, but then that seems to tedious

#### Trebla

$\bg_white k^2(2k^2-1) + (2k+1)^3\\= k^2(2k^2-1) + (2k+1)(4k^2+4k+1)\\= k^2(2k^2-1) + (2k+1)(2(k^2+2k+1)+ 2k^2 - 1)\\=(k^2+2k+1)(2k^2-1) + (k^2+2k+1)(4k+2)\\=(k+1)^2(2k^2+4k+1)\\=(k+1)^2(2(k+1)^2-1)$