Mathematical Induction

Question

Inequalities in M.I

Using the principle of mathematical induction show that 3ⁿ > n²

Answer 1

for n=0, 3^0 > 0^2, 1>0

n=1 3^1> 1^2, 3>1

n=2 3^2>2^2, 9>4

assume 3^k > k^2

try to show 3^(k+1) > (k+1)^2

3^k> k^2

multiply both sides by 3

3(3^k) > 3k^2

add the exponents of 3, 1+k

3^(k+1) >3k^2= 2K^2 + K^2 >K^2 + 2K +1 = (K+1)^2

cancel out the K^2 terms in 2K^2+K^2 > K^2 +2K +1

leaving 2K^2 > 2K + 1

divide by 2K

K> 1 + 1/K for K>2 1+1/K<2

QED for all natural numbers, n,

3^n > n^2