Prove by Mathematical induction that for any positive integer n≧10 , 2^n ＞ n^3

To prove 2^n > n^3.

Proof:

We pressume 2^n>n^3 for a particular n =r.

Then, we see whether it holds for r+1, i.e whether

2^(r+1) < (r+1)^3 is true given 2^r <  r^n

LHS :2^(r+1) = (2^2)*2

RHS= (r+1)^3 = r^3+3r^2+3r+1

But 3r^2+3r+1 < r^3  for all r>3.

Therefore, ( r+1)^3 < 2r^3 for all r >3 from which it  follows that

2^(r+1)  = 2*2^r  >  2r^3 > (r+1)^3  is true if and only if  2^r > r^3  for any k >3 .............................(1)

But 2^k is not greater than k^3  for  0<k<10.

But when k =10,  LHS 2^k = 1024 , RHS 10^3 =1000 . So, the relation, 2^k >  k^3 for k = 10 is true.

Therefore. 2^11 = 2(2^10) > 2*10^3 > 11^3 holds by (1).

So applying successively, the reation holds, for all r > 10.