Let P(n) be the statement: 2n ≥ 3n. If P(r) is true, show that P(r +

1.	Let P(n) be the statement: 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ϵ N?
Answer» If P(r) is true then 2^r ≥ 3r For, P(r+1) 2^r+1= 2.2^r For, x>3, 2x>x+3 So, 2.2^r > 2^r+ 3 for r >1 ⇒ 2^r+1>2^r+3 for r>1 ⇒ 2^r+1> 3r +3 for r>1 ⇒ 2^r+1> 3(r+1) for r>1 So, if P(r) is true, then P(r+1) is also true. For, n =1, P(1): L.H.S = 2 R.H.S = 3 As L.H.S < R.H.S So, it is not true for n = 1 Hence, P(n) is not true for all natural numbers.

Let P(n) be the statement: 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ϵ N?

Answer»

If P(r) is true then 2^r ≥ 3r

For, P(r+1)

2^r+1= 2.2^r

For, x>3, 2x>x+3

So, 2.2^r > 2^r+ 3 for r >1

⇒ 2^r+1>2^r+3 for r>1

⇒ 2^r+1> 3r +3 for r>1

⇒ 2^r+1> 3(r+1) for r>1

So, if P(r) is true, then P(r+1) is also true.

For, n =1, P(1):

L.H.S = 2

R.H.S = 3

As L.H.S < R.H.S

So, it is not true for n = 1

Hence, P(n) is not true for all natural numbers.