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Power by using the principle of Mathematical Induction that `n!>= 2^(n-1)` for every...

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roshan-rox | Valedictorian

Posted September 16, 2013 at 3:54 AM via web

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Power by using the principle of Mathematical Induction that `n!>= 2^(n-1)` for every positive integer n.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 16, 2013 at 4:15 AM (Answer #1)

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`n! >= 2^(n-1) `

`n = 1`

`LHS = 1! =1`

`RHS = 2^(1-1) = 1`

`LHS = RHS`

For `n = 1` result is true.

Let us assume that for n = p where `p>=1` and a positive integer the result is true.

`p!>=2^(p-1)`

`n = p+1`

`p!>=2^(p-1)` multiply both sides by `(p+1` ).

`p!xx(p+1)>=2^(p-1)xx(p+1)`

`(p+1)!>= 2^(p-1)xx(p+1)`

`p>=1` therefore `(p+1)>=2`

Therefore;

`(p+1)!>= 2^(p-1)xx2` is valid

`(p+1)!>= 2^(p+1-1)`

So for n = p+1 the result is true.

So from mathematical induction it is proved that `n! >= 2^(n-1) `

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