Induction Inequality Proofs (3 of 4: Introducing & transforming the inequality)

[沈博智-xy]

At 1:56, the other way could be to directly substituting in k>4

So:
LHS > k^2 - 2k - 1

> 4^2 - 2(4) - 1 {since k > 4}
= 16 - 8 - 1
= 7
> 0 {as required}

Therefore, 2^(k+1) > (k+1)^2
Therefore, the statement is true for n = k+1

Therefore, by the PMI, the statement is true for all n>4.

[gabrielmillee]

Why math is a better question. I'm not sure if I can comprehend anything beyond my fingers. But hes helped me first getting my GED, second in college.