1. ## Induction Inequality

Prove by Mathematical Induction that $\displaystyle n^2-11n+30 \geq 0$ for $\displaystyle n\geq1$.

Step 1: Prove true for n = 1
$\displaystyle 1^2-11(1)+30=20$
hence true for n = 1

Step 2: Assume true for n = k
$\displaystyle k^2-11k+30 \geq 0$

Step 3: Prove true for n = k + 1
$\displaystyle (k+1)^2-11(k+1)+30 \geq 0$

LHS = $\displaystyle k^2+2k+1-11k-11+30$

not really sure what to do from here, tried this approach

$\displaystyle k^2-11k+30+2k-10$
since from step 2 we know that $\displaystyle k^2-11k+30 \geq 0$, but not sure if this is right as I do not know what to do with the 2k - 10

any insight appreciated

2. Originally Posted by jacs
Prove by Mathematical Induction that $\displaystyle n^2-11n+30 \geq 0$ for $\displaystyle n\geq1$.

Step 1: Prove true for n = 1
$\displaystyle 1^2-11(1)+30=20$
hence true for n = 1

Step 2: Assume true for n = k
$\displaystyle k^2-11k+30 \geq 0$

Step 3: Prove true for n = k + 1
$\displaystyle (k+1)^2-11(k+1)+30 \geq 0$

LHS = $\displaystyle k^2+2k+1-11k-11+30$

not really sure what to do from here, tried this approach

$\displaystyle k^2-11k+30+2k-10$
since from step 2 we know that $\displaystyle k^2-11k+30 \geq 0$, but not sure if this is right as I do not know what to do with the 2k - 10

any insight appreciated
Note that $\displaystyle 2k-10\ge 0$ for all $\displaystyle k\ge 5$. You can check n= 1, 2, 3, and 4 "by hand", then start your induction at 5.

3. This is clearly not true. This is a quadratic that has a minimum turning point.

Note that $\displaystyle \displaystyle n^2 - 11n + 30 = (n - 5)(n - 6)$.

So the roots are $\displaystyle n = 5$ and $\displaystyle n = 6$.

That means that the function is $\displaystyle < 0$ for $\displaystyle 5 < n < 6$ and $\displaystyle \geq 0$ everywhere else.

4. thanks for that,
Wasnt sure if you could use that type of approach, or if it was thought of as a sort of cheat way around.

I did try $\displaystyle k^2-9k+20$

using the discriminant : $\displaystyle \Delta=(-9)^2-4(1)(20)$
= 1

but that proves that it is not positive definite (which is what i would have expected), it has two distinct rational roots, so it would NOT be larger than or equal to 0 for all k since it is indefinite, unless something is dodgy with my logic here

5. Originally Posted by Prove It
This is clearly not true. This is a quadratic that has a minimum turning point.

Note that $\displaystyle \displaystyle n^2 - 11n + 30 = (n - 5)(n - 6)$.

So the roots are $\displaystyle n = 5$ and $\displaystyle n = 6$.

That means that the function is $\displaystyle < 0$ for $\displaystyle 5 < n < 6$ and $\displaystyle \geq 0$ everywhere else.
your reply not posted yet when i did my last post, never even thought of checking that, duh, the question is set up dodgy then?
My discriminant argument confirms what you are saying?