Hi all!

 

i can't work out the method to solving this problem or proving it i suppose

 

i guess induction doesnt work but i am stumped for ideas, if anyone has any hints they could give that would be great!

 

i don't to be told exactly how to do it at the moment, but i do need a few hints to get me started.

 

Cheers

 

Sarah

ok i have come up with a solution....but it seems very iffy to me....anyways here it is....

_______________________________________________________________________

 

a(n) = 2*3^n - 3*2^n

a(n+1) = 2*3^{n+1) - 3*2^{n+1)

a(n-1) = 2*3^{n-1) - 3*2^{n-1)

 

6*a(n-1) = 12*3^{n-1} - 18*2^{n-1}

 

using the formula given (for a(n+1) ), rearrange that in terms of a(n):

i.e.

a(n) = (1/5)(a(n+1) + 6*a(n-1))

= (1/5)(2*3^{n+1} - 3*2^{n+1} + 12*3^{n-1} - 18*2^{n-1})

= (1/5)(2*3^{n+1} + 12*3^{n-1} - (3*2^{n+1} + 18*2^{n-1}))

= (1/5)(6*3^n+4*3^n - (6*2^n + 9*2^n))

= (1/5)(10*3^n - 15*2^n)

= 2*3^n - 3*2^n

= a(n)

 

Q.E.D

 

is that completely wrong? lol

i don't see how that would explain for n>=1 or n>=2 though....

 

anyways thats it

 

-Sarah

Are you sure you have not assumed what you were going to prove?

 

Anyway, a method that could be used is induction: First show that formula is correct for n = 1 and for n = 2, then show that if the formula is true for n = k and for n = k + 1, then it also is true for n = k + 2.

 

A more direct approach which does not assume that you already know or have guessed the formula for a_n is to use some knowledge of difference equations: a_(n + 1) = 5a_n - 6a_(n - 1) has the characteristic equation x^2 - 5x + 6 = (x - 2)(x - 3), so therefore a_n = C*2^N + d*3^N for alle n. We find C and D by using a_1 = 2C + 3D = 0 and a_2 = 4C + 9D = 2(2C + 3D) + 3D = 6. This gives D = 2 and C = -3, or a_n = 2*3^n - 3*2^n.

"Are you sure you have not assumed what you were going to prove?"

 

that is what i was afraid of...but i thought of it as if i assumed that was true then it would have to work when put into the given formula for a(n+1)...

 

is that kind of logic incorrect?

 

"Anyway, a method that could be used is induction: First show that formula is correct for n = 1 and for n = 2, then show that if the formula is true for n = k and for n = k + 1, then it also is true for n = k + 2."

 

yeah thats induction but i thought the step you assume in induction (true for n=k) is what we are trying to prove....

 

anyways i dunno, i am most likely wrong

yes that kind of logic is incorrect, though here it is largely a matter of presentation and gettign things in the right order.

 

induction assumes the truth of the statements for 1,..,k and uses these to show that k+1 is true. you just showed that certain things satisfied certain relationships. it may be rewritable to correct this but the first thing you did was to declare that a_{n+1} was what we wanted to show it was.

 

to correct it, assume it for a_{n-1} and a_{n} and then use the relation ot work out what a+{n+1 is, and tehn say this "is as required", ie don't state what a_{n+1} is before hand.

 

it remains to check, in this case the first two cases a_1 and a_2 since we are using the "previous two" to deduce the next.

ok would i also be able to do it by induction? (as below...)

 

show for n = 1

2*3^{1} - 3*2^{1} = 6 - 6 = 0

TRUE

 

show for n = 2

2*3^{2} - 3*2^{2} = 18 - 12 = 6

TRUE

 

assume true for n = k & n = k - 1

i.e.

a(k) = 2*3^{k} - 3*2^{k}

a(k-1) = 2*3^{k-1} - 3*2^{k-1}

 

by defn.

a(n+1) = 5*a(n) - 6*a(n-1)

or

a(k+1) = 5*a(k) - 6*a(k-1)

 

.....then substitute in the assumption from above and eventually it should give :

a(k+1) = 2*3^{k+1} - 3*2^{k+1}

 

 

hows that? is that correct now? that method made more sense to me

 

Cheers

 

Sarah

Yes, that's correct. In fact, between "substituting" and "eventually" is hardly one step.

"In fact, between "substituting" and "eventually" is hardly one step."

 

what do you mean?

He means this:

 

a(k+1) = 5*a(k) - 6*a(k-1)

 

.....then substitute in the assumption from above and eventually it should give :

a(k+1) = 2*3^{k+1} - 3*2^{k+1}

 

You obviously need to show more working in between those two steps

lol yeah i see well it would have taken ages to type out, but if i explained what the working was,.... so the proof is still good right?

Yeah, that gap just needs to be filled in.

do'nt use "should" as that implies that you haven't actually done the work and think it probably is true. this is maths, yo don't do that.

no no i have done the work, but as it takes ages to fill in that bit (to type into here i mean) i havent put it up.

thanks for you help everyone!