induction help~ (1 Viewer)

[shkspeare]

i need help with this question :S

 

[shkspeare]

any1?

 

[shkspeare]

thank u

 

[Fosweb]

Scanned and attached

 

[shkspeare]

thanks fos ur a lifesaver.. i know where i went wrong in my working now

 

[shkspeare]

ok for this question

2^n > n^2

some1 check my working i dont know how to answer

for n = 1
LHS = 2^1 = 2
RHS = 1^2 = 1
LHS > RHS
Therefor true for n =1

Assume n=k
2^k > k^2
for n = k+1
2^(k+1) > (k+1)^2
LHS = 2(2^k)
> 2(k^2) (from 2^k = k^2)

whats next ???!!!

 

[Fosweb]

Firstly this only works when n>4, so you can't start with n=1 (whilst it does work for 1, it wont for 2, 3, 4)

Theres a trick in here.
You have shown that 2^k > k²
Which means that 2.2^k > 2.k² (Just multiplied by 2 on each side)


So when you are proving for n = k+1:

LHS = 2^k+1 = 2.2^k > 2.k²

Now all you have to do is prove that 2.k² > (k+1)² and you have proven what you need. (Remember for when you prove this that k>4)

And thats it!

 

[shkspeare]

woww thanks a lot again

 

[Fosweb]

No problems.

This is probably about as hard as 3U induction will get (maybe a bit longer, but with the same style of 'tricks')

 

[sven0023]

if you are having trouble on these questions you can always
do expand LHS, Expand RHS etc

Ie LHS =
=
=
=

RHS =
=
=


LHS = RHS .:. true for n = k + 1