Question

If P(n) is the statement 2^n >3n and if P(k) is true, show that P(k+1) is also true.​

Answer 1

$\underline{\underline{\huge{\pink{\tt{\textbf Answer :-}}}}}$

Given as 

P (n) = “2n ≥ 3n” and p(r) is true.

We have, P (n) = 2n ≥ 3n

Here, P (r) is true

Therefore,

2r ≥ 3r

Then, let’s multiply both sides by 2

2 × 2r ≥ 3r × 2

2r + 1 ≥ 6r

2r + 1 ≥ 3r + 3r [since 3r >3 = 3r + 3r ≥ 3 + 3r]

∴ 2r + 1 ≥ 3(r + 1)

Thus, P (r + 1) is true.