2^(n-1) = 2x2x2x2x..........2
............|__(n-1)products___|
n! = 1x2x3x4x5x.....n = 2x3x4x....n [a total of n-1 product terms]
.'. 2x2x2x2x..........(n-1)times ≤ 2x3x4x5x........(n)
2^(n-1)=(n)!, only when n=1, LHS=RHS=1
@samwise, i see he prefers mathematical induction, but mine is a simpler method, unless mentioned mathematical induction!!!!!
2^(0-1) = 1/2 and 0! = 1 > 1/2, so it holds for n=0.
2^(1-1) = 1 and 1! = 1, so it holds for n=1.
Suppose the assertion holds for some n=k where k>1. That is:
2^(k-1) ≤ k!
Then it must hold for n=k+1 as well, because
2^[(k+1)-1] = 2^k = 2 * 2^(k-1)
and
(k+1)! = (k+1) * k!
2 ≤ k+1
Since we know it holds for n=1, this proves it holds for any integer value of n>1 as well, by mathematical induction.
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2^(n-1) = 2x2x2x2x..........2
............|__(n-1)products___|
n! = 1x2x3x4x5x.....n = 2x3x4x....n [a total of n-1 product terms]
.'. 2x2x2x2x..........(n-1)times ≤ 2x3x4x5x........(n)
2^(n-1)=(n)!, only when n=1, LHS=RHS=1
@samwise, i see he prefers mathematical induction, but mine is a simpler method, unless mentioned mathematical induction!!!!!
2^(0-1) = 1/2 and 0! = 1 > 1/2, so it holds for n=0.
2^(1-1) = 1 and 1! = 1, so it holds for n=1.
Suppose the assertion holds for some n=k where k>1. That is:
2^(k-1) ≤ k!
Then it must hold for n=k+1 as well, because
2^[(k+1)-1] = 2^k = 2 * 2^(k-1)
and
(k+1)! = (k+1) * k!
and
2 ≤ k+1
Since we know it holds for n=1, this proves it holds for any integer value of n>1 as well, by mathematical induction.