Question

please help me to solve this question step by step​

Answer 1

Answer:

see below

Step-by-step explanation:

proof : by mathematical induction

P(n) : n ! ≥ 2^ ( n - 1 )

now.....

for n= 1 ; 1 = 2^ ( 1 - 1 ) = 2^0 = 1

hence....P(1) is true

let for n = k > 2; P(k) be also true

i,e P(k) : k ! > 2^ ( k - 1 )

now...

k ! (k+1) ! > 2^ ( k - 1 ) (k+1)

but...

2^ ( k - 1 ) (k+1) < 2^ ( k - 1 ) × 2 ( bcoz k > 2)

k ! (k+1) > 2^ ( k - 1 ) × 2

( k + 1 ) ! > 2^ ( k - 1 + 1 )

( k + 1 ) ! > 2^ [ ( k +1) - 1 ]

therefore....

for n = k + 1 ; P ( k + 1 ) also true

hence P ( n ) is true true for all n

Answer 2

Answer:

By mathematical induction

P(n) : n ! ≥ 2^ ( n - 1 )

now.....

for n= 1 ; 1 = 2^ ( 1 - 1 ) = 2^0 = 1

hence....P(1) is true

let for n = k > 2; P(k) be also true

i,e P(k) : k ! > 2^ ( k - 1 )

now...

k ! (k+1) ! > 2^ ( k - 1 ) (k+1)

but...

2^ ( k - 1 ) (k+1) < 2^ ( k - 1 ) × 2 ( bcoz k > 2)

k ! (k+1) > 2^ ( k - 1 ) × 2

( k + 1 ) ! > 2^ ( k - 1 + 1 )

( k + 1 ) ! > 2^ [ ( k +1) - 1 ]

therefore....

for n = k + 1 ; P ( k + 1 ) also true

hence P ( n ) is true true for all n

Step-by-step explanation:

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