You need to use mathematical induction to prove the inequality, hence, you need to perform the following two steps, such that:
Step 1: Basis: Prove that the statement holds for n = 1
`(1+a)^1 >= 1*a => 1 + a > a`
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds.
`P(k): (1+a)^k >= k*a` holds
`P(k+1): (1+a)^(k+1) >= (k+1)*a`
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side of inequality such that:
`(1+a)^(k+1) = (1+a)^k*(1+a) >= k*a*(1+a) >= (k+1)*a`
Opening the brackets yields:
`ka + ka^2 >= ka + a`
Notice that `ka^2 > a` , hence, the inequality `ka + ka^2 >= ka + a` holds.
Hence, since both the basis and the inductive step hold, the statement `P(n): (1+a)^n >= n*a` holds for all indicated values of n.
No comments:
Post a Comment