Question

use the binomial theorem to show that

2^n > n(n-1)(n-2)/3!

true for all n in natural numbers

Answer 1

Couple of tricks here. Let's take the right hand side
\[\frac{n(n-1)(n-2)}{3!}\]
Multiply both numerator and denominator by (n-3)
\[\frac{n(n-1)(n-2)(n-3)!}{(n-3)!3!}\]
The numerator clearly becomes n!
Thus, the RHS can be written as :
\[{n}\choose3\]

Now, write 2^n as
\[(1+1)^n\]
If you expand it, it will look like:
\[{n \choose 0} 1^n + {n \choose 1}1^{(n-1)}1^1 + {n \choose 2}1^{n-2}1^2 + {n \choose 3} 1^{n-3}1^3 + ...\]

The fourth term in this expansion is basically the RHS.

So, by this manipulation, we find that 2^n = RHS + some positive terms (cause n is natural), hence we prove the given inequality.

Answer 2

:O thats like so easy.........haha thanks i totally missed that answer on my exam then