OpenStudy (anonymous):

I need help with proof by induction, I don't quite get it.

OpenStudy (anonymous):

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = [n(6k^2 - 3n - 1)]/2

OpenStudy (anonymous):

I already proved that it works for 1, and have it as n=k, I'm just having trouble proving k+1

OpenStudy (anonymous):

$1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = [n(6k^2 - 3n - 1)]/2$ ... you can use \ [ and \ ], so it's easier to read. Opposite click on mine to see how it works.

OpenStudy (anonymous):

If n=k, the Inductive Hypothesis, what do you get?

OpenStudy (anonymous):

This is my equation for k+1, the first fraction is my value for n=k Its mostly the algebra I need help with $\frac{ k(6k ^{2} - 3k - 1) }{ 2 } + \frac{ (3(k+1) - 2) }{ 1 }$

OpenStudy (dtan5457):

Essentially all you need to do is plug n+1 into the left side formula, and add that with the total sum formula (the formula on the right side). Then set that equal to (n+1) plugged into the right side. so (3(n+1)-2)^2+[n(6k^2 - 3n - 1)]/2=[n+1)(6(n+1)^2 - 3(n+1) - 1)]/2

OpenStudy (dtan5457):

im assuming the k was suppose to be an n

OpenStudy (anonymous):

I was taught to write the equation as n=k

OpenStudy (anonymous):

how do I show that it equals that?

OpenStudy (anonymous):

or can it just be shown as a one step thing?

OpenStudy (anonymous):

Your almost done, I think, use your inductive hype. to rewrite the first part of your expression to yeild $1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 + \frac{ (3(k+1) - 2) }{ 1 }$

OpenStudy (anonymous):

Maybe there is an algebra error, I'm not getting $\frac{3(k+1)-2}{1}\overset{?}{=} (3k-1)^2$

OpenStudy (anonymous):

$\frac{ (k+1)\left(6(k+1) ^{2} - 3(k+1) - 1\right) }{ 2 }$

OpenStudy (anonymous):

so, between $\frac{ k(6k ^{2} - 3k - 1) }{ 2 }+(3k-1)^2$ and $\frac{ (k+1)\left(6(k+1) ^{2} - 3(k+1) - 1\right) }{ 2 }$ there is just one step?

OpenStudy (anonymous):

nope, lots of algebra I think...

OpenStudy (anonymous):

ok, that's what I really need help with

OpenStudy (anonymous):

Maybe totally expand both LHS and RHS, might not be fastest, but it will work. $RHS=\frac{ (k+1)\left(6(k+1) ^{2} - 3(k+1) - 1\right) }{ 2 }$ $LHS=\frac{ k(6k ^{2} - 3k - 1) }{ 2 }+(3k-1)^2$

OpenStudy (anonymous):

ok, thanks, I'll see what I can do

OpenStudy (anonymous):

Something is wonky, i get $RHS = 1 + (11 k)/2 + (15 k^2)/2 + 3 k^3$ $LHS = 1 - 7 k + (15 k^2)/2 + 3 k^3$

OpenStudy (anonymous):

I think I may have something figured out. Thanks for your help

OpenStudy (anonymous):

np... I made an error, the LHS is $\frac{ k(6k ^{2} - 3k - 1) }{ 2 }+(3k+1)^2$