Question

2^n>n^2

Answer 1

induction

Answer 2

where n>4

Answer 3

\[2^5=32>5^2=25\]
\[2^k>k^2\]
\[2^{k+1}>(k+1)^2\]
\[2(2^k)>2k^2\]inductive step
but I cant move from here\[2k^2>(k+1)^2\]

Answer 4

@phi how do I show this is true

Answer 5

\[2k^2>k^2+2k+1\]

Answer 6

If we assume true for k
2^k > k^2

2* 2^k > 2 k^2
2^(k+1) > k^2 + k^2

if we can prove that k^2 > 2k+1 then we can say
2^(k+1) > k^2 + 2k +1
2^(k+1) > (k+1)^2

so you need to prove k^2 > 2k+1

Answer 7

okay can we use the fact that n>4
so
k>4
k^2>16
2k+1>9

Answer 8

if you don't mind you can restart with your own method

Answer 9

I would use induction to prove k^2 > 2k+1
once you have that fact you can say
2^(k+1) > k^2 + k^2 > k^2 + 2k+1
2^(k+1) > k^2 + 2k+1
2^(k+1) > (k+1)^2

Answer 10

just a quick question... I realise this is mathematics induction.... what about

2^4 = 4^2

just a thought

Answer 11

2^4 = 4^2
probably why they require n>4

Answer 12

yes the condition saysif n is a positive integer >4

Answer 13

can you help me with a new one

Answer 14

here is induction for k^2 > 2k+1 , k>4
base case: k=5 25>11 true
assume true for k
k^2 > 2k+1
k^2 + 2k + 1 > 2k+1 + 2k +1
(k+1)^2 > 2k+2 + 2k
(k+1)^2 > 2(k+1) + 2k > 2(k+1) +1 (clearly true for k>4)

(k+1)^2 > 2(k+1) +1 is true for k+1, and therefore true in general

Answer 15

wonderful thanks ,so now we can conclude the previuos one