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

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- Dec 3rd 2012, 02:47 PMrchrist3PRoof by induction: Inequality
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 - Dec 3rd 2012, 04:18 PMcoolgeRe: PRoof by induction: Inequality
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.) - Dec 3rd 2012, 06:56 PMrchrist3Re: PRoof by induction: Inequality
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? - Dec 4th 2012, 06:06 PMcoolgeRe: PRoof by induction: Inequality
Your reasoning is correct.