sequence problem

[Sarahisme]

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

[Sarahisme]

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

[Algebracus]

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.

[Sarahisme]

"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

[matt grime]

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.

[Sarahisme]

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

[DQW]

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

[Sarahisme]

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

 

what do you mean?

[Dave]

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

[Sarahisme]

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?

[Dave]

Yeah, that gap just needs to be filled in.

[matt grime]

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.

[Sarahisme]

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.

[Sarahisme]

thanks for you help everyone!