induction proof

Question

proof that for n>=2 it is true that:

2^(1+2+3+...+n)>n!

Answer 1

First we must show that 2^k > k for any integer k.

for k=0 ---> 2^0 = 1 > 0

for k=1 ---> 2^1 = 2 > 1

the given induction hypothesis is 2^k > k for integer k>1

2^(k+1) = 2^k * 2 > k*2 = k+k > k+1

Therefore 2^n > n for any integer n

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For n=2: 2^(1+2) = 2^3 = 8 > 2=2!

For n=3: 2^(1+2+3) = 2^(6) = 64 > 6 = 3!

The given induction hypothesis is 2^(1+2+3+....+n) > n!

statement reason

2^(1+2+3+....+n+(n+1)) =

2^(1+2+3+....+n)*2^(n+1) > property of exponent

n!*2^(n+1) > substitutes induction hypothesis

n! * (n+1) = previous theorem proven above

(n+1)!