Show that this sequence satisfies the recurrence relation

[bdh2991]

Homework Statement

Let d₀, d₁, d₂,... be defined by the formula d_n = 3ⁿ - 2ⁿ for all integers n ≥ 0. Show that this sequence satisfies the recurrence relation.

d_k = 5d_k-1 - 6d_k-2.

Homework Equations

The Attempt at a Solution

I found that d_k = 3^k - 2^k

d_k-1 = 3^k-1 - 2^k-1
d_k-2 = 3^k-2 - 2^k-2

after plugging d_k-1 and d_{k - 2} into the formula d_k = 5d_k-1 - 6d_k-2, i am stuck and do not understand how to do the algebra if that is what i 'm supposed to be doing...any help?

 

[SammyS]

Homework Statement

Let d₀, d₁, d₂,... be defined by the formula d_n = 3ⁿ - 2ⁿ for all integers n ≥ 0. Show that this sequence satisfies the recurrence relation.

d_k = 5d_k-1 - 6d_k-2.

Homework Equations

The Attempt at a Solution

I found that d_k = 3^k - 2^k

d_k-1 = 3^k-1 - 2^k-1
d_k-2 = 3^k-2 - 2^k-2

after plugging d_k-1 and d_{k - 2} into the formula d_k = 5d_k-1 - 6d_k-2, i am stuck and do not understand how to do the algebra if that is what i 'm supposed to be doing...any help?

Take

d_k-1 = 3^k-1 - 2^k-1

and

d_k-2 = 3^k-2 - 2^k-2 '​

Plug those into

5d_k-1 - 6d_k-2 .​

Do some algebra & see what you get.

 

[bdh2991]

Take

d_k-1 = 3^k-1 - 2^k-1

and

d_k-2 = 3^k-2 - 2^k-2 '​

Plug those into

5d_k-1 - 6d_k-2 .​

Do some algebra & see what you get.

This is what i tried doing at i stated above...after you plug them in i get

3^k - 2^k = 5( 3^k-1 - 2^k-1) - 6(3^k-2 - 2^k-2), I don't really understand what i can do with that algebraically...i must just be missing it...

 

[Dick]

This is what i tried doing at i stated above...after you plug them in i get

3^k - 2^k = 5( 3^k-1 - 2^k-1) - 6(3^k-2 - 2^k-2), I don't really understand what i can do with that algebraically...i must just be missing it...

Use 3^(k-1)=3*3^(k-2) and 2^(k-1)=2*2^(k-2).

 

[bdh2991]

Use 3^(k-1)=3*3^(k-2) and 2^(k-1)=2*2^(k-2).

ok so changing the equation to that gives me:

5(3*3^k-2 - 2*2^k-2) - 6( 3^k-2 - 2^k-2)

i can see how that gave me some like terms but multiplying through gives me:

15 * 3^k-2 - 10 * 2^k-2 - 6 * 3^k-2 - 6 * 2^k-2

i feel like that wasn't where you were leading me lol...i'm sorry, I'm not that great at algebra

 

[Dick]

ok so changing the equation to that gives me:

5(3*3^k-2 - 2*2^k-2) - 6( 3^k-2 - 2^k-2)

i can see how that gave me some like terms but multiplying through gives me:

15 * 3^k-2 - 10 * 2^k-2 - 6 * 3^k-2 - 6 * 2^k-2

i feel like that wasn't where you were leading me lol...i'm sorry, I'm not that great at algebra

I guess not. But you're almost there. Collect the 3^(k-2) terms. What do you get? And you've a sign error on the last term. Could you fix it?