uni maths help (1 Viewer)

[redruM]

i was having problems understanding what an example is doing. maybe because it is late, or maybe because i havent done any maths in 2 years.

anyways, see attatched. the part boxed in red is what i have trouble understanding. if someone could provide some sort of explanation, it would be great.

thanks in advance.

 

[redruM]

anyone...?

 

[maniacguy]

Hmm... long time since I've been here. Anyway:
3^{n+1}
> 3n^3 (since 3^{n+1} = 3*3^n > 3*n^3 as 3^n > n^3)
= n^3 + 2*n^3
> n^3 + 2*n^2*3 (since n>3 is given in this instance)
= n^3 + 6n^2
= n^3 + 3n^2 + 3n^2
> n^3 + 3n^2 + 6n (since n>3 means n>2)
> n^3 + 3n^2 + 3n + 1 (as 3n > 1)
= (n+1)^3

 

[who_loves_maths]

hi redruM,

i suspect the only place you do not understand are parts like:

2n^3 > 6n^2 , and 3n^2 > 6n , and 3n > 1 .

the reason for these steps is that your original assumption encompasses the number 'n' such that n: n > 3 , and not anything less than 3.

and so that's how you are able to say that for this specific case of induction: 2n^3 > 6n^2 ---> since 2n^3 = 2n(n^2), and when n > 3, then clearly 2n(n^2) > 2*3(n^2) = 6n^2 ; hence, you get 2n^3 > 6n^2 .

this is the same for the rest: 3n^2 > 6n, and 3n > 1, etc...