Question

Help with mathematical induction?

The problem is (3n-2)^2 = (n(6n^2 -3n -1))/2

I know that n=1 is okay. Both sides equal 1. But when I start plugging in k+1, I get lost. Can someone help me test k+1?

Answer 1

You are proving it with Induction.
This means you have to do the following:
1. Check to see if it is true for n=1. Done.
2. Suppose it is true for n=k, then prove it's true for n=k+1.

Just to be safe, 2. means:\[(3k+2)^2=\frac{ k(6k^2-3k-1) }{ 2 }\]is true.
Not let's plug in n=k+1.
Left side:\[(3(k+1)+2)^2\]
Right side:\[\frac{ (k+1)(6(k+1)^2-3(k+1)-1) }{ 2 }\]
I suppose you have come this far.

Answer 2

ok.... you need to find the term in (k + 1)

\[T_{k +1} = (3(k + 1) -2)^2 \]

and

the sum of k terms is

\[S_{k} = \frac{k(6k^2 -2k -1)}{2}\]

you need to show

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

Answer 3

so looking at

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

Answer 4

@ZeHanz,
thanks. Yes, I have gotten that far. It is what I have to do from there that I am confused.

@campbell_st ,
thanks. I think I understand what you mean so far.

Answer 5

which I think is

\[T_{k +1}+ S_{k} = \frac{ 6k^3 + 15k^2 + 11k + 2}{2}\]

Answer 6

now its expanding

\[S_{k + 1}\] to show its equal

Answer 7

Look at campbell_st's explanation...

Answer 8

the first part of the expansion is

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

which then becomes

\[S_{k + 1} = \frac{(k + 1)(6k^2 + 9k + 2)}{2}\]

it shoule be straight forward from here to prove for k + 1

I've done it and I know it works..

Answer 9

all you need to do is show

\[T_{k + 1} + S_{k} = S_{k + 1}\]

Answer 10

I see what you're doing and I'm trying to replicate it. I can't seem to find how tk+1 + sk = sk+1. But you said it was true right?

Answer 11

@campbell_st,
is this statement false?

Answer 12

the statement is true

this is what you are trying to prove

\[T_{k + 1} + S_{k} = S_{k + 1} \]

so if you know the sum to a specific term and then add the next term you should get the sum to the k + 1 th term

so

\[\frac{6k^3 + 15k^2 + 11k + 2}{2} = \frac{(k + 1)(6k^2 + 9k + 2)}{2} \]

distribute on the right hand side gives

\[\frac{6k^3 + 15k^2 + 11k + 2}{2} = \frac{ 6k^3 + 9k^2 + 2k + 6k^2 + 9k + 2}{2}\]

collect the like terms on the right and you'll see that boths sides are equal, as required.

Answer 13

hope this make sense.

Answer 14

OHHHHH I see. Thanks so much man. I actually understand this now. :P