Question

for every integer n, n^2≥n. prove by cases

Answer 1

step  by step

You proved it's true for n=5. Now suppose it's true for some integer n≥5. The aim is to prove it's true for n+1. But

(♠)2n+1=2×2n>2×n2>(n+1)2.

The first inequality follows from the induction hypothesis and as for the second, we know that (n−1)2≥42>2, since n≥5. We can expand this inequality (n−1)2>2 as follows:

n2−2n+1>n2−2n−1>2n2−2n−1>2n2>20n2n2+2n+1=(n+1)2,