Sequences that satisfay the same recurrence relation

[Jim01]

Homework Statement

Let a₀, a₁, a₂..., be defined by the formula a_n = 3n + 1, for all integers n >= 0. Show that this sequence satisfies the recurrence relation a_k = a_k-1 + 3, for all integers k >=1.

Homework Equations

for all integers n >= 0, a_n = 3n + 1

for all integers k >= 1, a_k = a_k-1 + 3

The Attempt at a Solution

I have no idea how to proceed. In the one example given us in the book, we are given the initial conditions for each sequence (a₁ = 2 and b₁ = 1) and the formulas are exactly the same except that one is a_k = 3a_k-1 and the other is b_k = 3b_k-1. I am unable to relate the example in thge book to the question.

I have looked on youtube but can only find videos on how to compute terms of a recursively defined sequence, which I know how to do.

 

[tiny-tim]

Hi Jim01!

If a_n = 3n + 1,

what is a_n-1 ?

And so what is a_n - a_n-1 ?

 

[Jim01]

Hi Jim01!

If a_n = 3n + 1,

what is a_n-1 ?

And so what is a_n - a_n-1 ?

If I am understanding you correctly, then

If a_n = 3n + 1,

then

a_n-1 = 3(n - 1) + 1
= 3n - 3 + 1
= 3n - 2

therefore

a_n - a_n-1 = 3n + 1 - 3n - 2
= -1

I'm sorry for being so thick-headed but I don't know what this is telling me.

 

[Jim01]

If I am understanding you correctly, then

If a_n = 3n + 1,

then

a_n-1 = 3(n - 1) + 1
= 3n - 3 + 1
= 3n - 2

therefore

a_n - a_n-1 = 3n + 1 - 3n - 2
= -1

I'm sorry for being so thick-headed but I don't know what this is telling me.

OK. I think I may get it. The first thing you did was to make all the subscripts the same, so that a₀, a₁, a₃, ... is defined by a_k = 3k + 1 and a_k-1 = 3(k - 1) + 1. I didn't know you could do that, although since the letters are arbitrary, it makes sense.

You then input the answer to a_k-1 into the the formula a_k = a_k-1 + 3.

so a_k-1 + 3
= 3(k - 1) + 1 + 3
= 3k - 3 + 1 + 3
= 3k + 1
= a_k

Ok. I got it. Thank you very much.

 

[HallsofIvy]

If I am understanding you correctly, then

If a_n = 3n + 1,

then

a_n-1 = 3(n - 1) + 1
= 3n - 3 + 1
= 3n - 2

therefore

a_n - a_n-1 = 3n + 1 - 3n - 2
= -1

Your subtraction is wrong. You want 3n+1- (3n-2)= 3n+1- 3n+ 2= 3, not -1.

I'm sorry for being so thick-headed but I don't know what this is telling me.

 

[Jim01]

Your subtraction is wrong. You want 3n+1- (3n-2)= 3n+1- 3n+ 2= 3, not -1.

You are absolutely correct. Thank you for pointing that out to me.