Question

Show that n^3>(N+1)^2 for all natural numbers n>=3.

Answer 1

So I showed that P(3) is true and I changed P(n) to P(k) so that I could prove P(k+1). I added 3K^2+3k+1 to both sides to get (k+1)^3>3k^2+5k+2 but I'm not sure what's next.

Answer 2

Where I'm up to:\[(k+1)^3>(k^2+2k+1)+(3k^2+3k+1) \]

Answer 3

i wanna use induction?

Answer 4

I do too. Go ahead.

Answer 5

expand

Answer 6

\[(k+1)^3\]\[=k^3+3k^2+3k+1\]\[>(k+1)^2+3k^2+3k+1\]\[=k^2+2k+1+3k^2+3k+1\]\[=4k^2+5k+1\]\[=\text{stuck xD}\]

Answer 7

alternatively, it should be clear that \[f(n)=n^3-(n+1)^2\] is increasing on \((3,\infty)\) and \[f(3)>0\]

Answer 8

can I say