If P(n) : n! > 2^{n – 1}, n ∈ N, then P(n) is true for all n > _____________.

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Solution

P(n) : n! > 2^{n – 1}; n ∈ N

for n = 1,
P(1) : 1! > 2^1–1
i.e 1 > 2° = 1
i.e 1 > 1
which is false a statement

for n = 2
P(2) : 2! > 2^2–1
i.e 2 > 2¹
i.e 2 > 2
which is again a false statement.
for n = 3
P(3) : 3! > 2^3–1
i.e 6 > 2²= 4
i.e 6 > 4 which is true

for n = 4
P(4) : 4! > 2^4–1
i.e 24 > 2³= 8 which is true

Hence, P(n) : n! > 2^{n – 1}is true
for n > 2

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