Mathematical Induction Algebra simplification? (1 Viewer)

[cchan334]

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]

(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]

(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]

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]

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]

Expanding and simplifying is the most efficient approach here. It’s not that bad lol.

If you insist on not expanding cubics but happy to expand quadratics then the following manipulation can be done, but it’s longer and trickier than fully expanding.