Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

This is my equation for k+1, the first fraction is my value for n=k
Its mostly the algebra I need help with

\[\frac{ k(6k ^{2} - 3k - 1) }{ 2 } + \frac{ (3(k+1) - 2) }{ 1 }\]

Answer 6

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

Answer 7

im assuming the k was suppose to be an n

Answer 8

I was taught to write the equation as n=k

Answer 9

how do I show that it equals that?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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?

Answer 15

nope, lots of algebra I think...

Answer 16

ok, that's what I really need help with

Answer 17

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

Answer 18

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

Answer 19

Something is wonky, i get

\[RHS = 1 + (11 k)/2 + (15 k^2)/2 + 3 k^3\]
\[LHS = 1 - 7 k + (15 k^2)/2 + 3 k^3\]

Answer 20

I think I may have something figured out.

Thanks for your help

Answer 21

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