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anonymous
 5 years ago
use the binomial theorem to show that
2^n > n(n1)(n2)/3!
true for all n in natural numbers
anonymous
 5 years ago
use the binomial theorem to show that 2^n > n(n1)(n2)/3! true for all n in natural numbers

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sid1729
 5 years ago
Best ResponseYou've already chosen the best response.0Couple of tricks here. Let's take the right hand side \[\frac{n(n1)(n2)}{3!}\] Multiply both numerator and denominator by (n3) \[\frac{n(n1)(n2)(n3)!}{(n3)!3!}\] 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: \[{n \choose 0} 1^n + {n \choose 1}1^{(n1)}1^1 + {n \choose 2}1^{n2}1^2 + {n \choose 3} 1^{n3}1^3 + ...\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0:O thats like so easy.........haha thanks i totally missed that answer on my exam then
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