• anonymous
use the binomial theorem to show that 2^n > n(n-1)(n-2)/3! true for all n in natural numbers
  • schrodinger
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  • sid1729
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.
  • anonymous
:O thats like so easy.........haha thanks i totally missed that answer on my exam then

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