I need to prove that for n>= 7, 3n^2>17n+10
I know the first steps proving for n=7, then assuming for k, but once i need to prove for (k+1) i am stumped.
Thank you for your time and help!
Rob
Base case: For n = 7, you have
3 n^{2} = 3 * 49 = 147
17n + 10 = 17 * 7 + 10 = 129
So, 3n^{2}>17n+10
Induction: Assume the inequality holds for n = k > 7. That is, assume that 3k^{2}>17k+10. We need to prove that the inequality holds for n = k +1. That is, we need to prove that 3(k+1)^{2}>17(k+1)+10.
But 3(k+1)^{2} = 3k^{2} + 6k + 3. We know from our assumption 3k^{2}>17k+10. So, we get 3(k+1)^{2} = 3k^{2} + 6k + 3 > 17k + 10 + 6k + 3 > 17k + 17 + 10 = 17(k+1) + 10. (Note: 6k+3 > 17 when k > 7.)
thank you for your help! now i worked it all out and it went like this:
17k+10+6k+3>17k+17+10
: 23K+13>17k+27
: 23k>17k+14
:6k>14
: k> 2.3
i feel like i did something wrong... or this is right because if it works for k>2.3 we know it is true bc out k is >= 7?