Sequences that satisfay the same recurrence relation

AI Thread Summary
The sequence defined by an = 3n + 1 satisfies the recurrence relation ak = ak-1 + 3 for k >= 1. To demonstrate this, the relationship between an and an-1 is established, where an-1 = 3(n - 1) + 1. By substituting an-1 into the recurrence relation, it simplifies to ak = 3k + 1, confirming the sequence's validity. A common mistake in subtraction was corrected during the discussion, leading to the realization that an - an-1 equals 3, not -1. This clarification helped solidify the understanding of the recurrence relation.
Jim01
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Homework Statement



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



Homework Equations



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

for all integers k >= 1, ak = ak-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 ak = 3ak-1 and the other is bk = 3bk-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.
 
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Hi Jim01! :smile:

If an = 3n + 1,

what is an-1 ?

And so what is an - an-1 ? :wink:
 
tiny-tim said:
Hi Jim01! :smile:

If an = 3n + 1,

what is an-1 ?

And so what is an - an-1 ? :wink:

If I am understanding you correctly, then

If an = 3n + 1,

then

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

therefore

an - an-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 said:
If I am understanding you correctly, then

If an = 3n + 1,

then

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

therefore

an - an-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 ak = 3k + 1 and ak-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 ak-1 into the the formula ak = ak-1 + 3.

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

Ok. I got it. Thank you very much.
 
Jim01 said:
If I am understanding you correctly, then

If an = 3n + 1,

then

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

therefore

an - an-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.
 
HallsofIvy said:
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.
 
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