# Sequence and recurrence relationships

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#1
How would you use algebra to prove that both sequences have the same numerical value for all values of n?
Last edited by lavely; 1 month ago
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1 month ago
#2
A variety of ways and Im not 100% sure which one they want. But you could try writing down u_n and u_(n+1) in terms of n using A, then showing that you get the recurrence relationship B when you combine them. You'd also need to "show" u_1 = 0 using A.
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
A variety of ways and Im not 100% sure which one they want. But you could try writing down u_n and u_(n+1) in terms of n using A, then showing that you get the recurrence relationship B when you combine them. You'd also need to "show" u_1 = 0 using A.
Thanks, I’ve tried the first step you recommended, could you try an explain the bit of the solution highlighted in yellow?

I’ve looked at it for ages and I can’t work out what they have done
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1 month ago
#4
They basically do the opposite of what I was suggesting, i.e. using B to get A (rather than using A to get B). So the aim is to get the closed form (position to term or n -> u_n) using the recurrence relationship in sequence B. I think what I proposed is simpler, but both are valid.

So iteratively / recursively working backwards using sequence B you get
u_n = u_(n-1) + 2 = u_(n-2) + 2 + 2 = u_(n-3) + 2 + 2 + 2 = ... = u_1 + 2(n-1) = 2n - 2
Which is sequence A description.
Last edited by mqb2766; 1 month ago
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#5
(Original post by mqb2766)
They basically do the opposite of what I was suggesting, i.e. using B to get A (rather than using A to get B). So the aim is to get the closed form (position to term or n -> u_n) using the recurrence relationship in sequence B. I think what I proposed is simpler, but both are valid.

So iteratively / recursively working backwards using sequence B you get
u_n = u_(n-1) + 2 = u_(n-2) + 2 + 2 = u_(n-3) + 2 + 2 + 2 = ... = u_1 + 2(n-1) = 2n - 2
Which is sequence A description
I'm really so sorry, I understand what you are trying to say, but I still don't understand how to do the iterative/recursive part you have shown. Could you explain how you work backwards from sequence B? Your help is greatly appreciated!!
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1 month ago
#6
(Original post by lavely)
I'm really so sorry, I understand what you are trying to say, but I still don't understand how to do the iterative/recursive part you have shown. Could you explain how you work backwards from sequence B? Your help is greatly appreciated!!
Using B, you interpret the recurrence relationship as
current value = previous value + 2
and keep iterating (current -> previous and calculate new previous).

So substituting different values for the current value (n, n-1, n-2, ..., 2) gives:
u_n = u_(n-1) + 2
u_(n-1) = u_(n-2) + 2
u_(n-2) = u_(n-3) + 2
...
u_3 = u_2 + 2
u_2 = u_1 + 2
u_1 = 0
Then you just substitute back up the chain to get
u_2 = 0 + 2 = 2
u_3 = 2 + 2 = 2*2
u_4 = 2*2 + 2 = 3*2
...
u_n = (n-1)*2
Last edited by mqb2766; 1 month ago
0
#7
(Original post by mqb2766)
Using B, you interpret the recurrence relationship as
current value = previous value + 2
and keep iterating (current -> previous and calculate new previous).

So substituting different values for the current value (n, n-1, n-2, ..., 2) gives:
u_n = u_(n-1) + 2
u_(n-1) = u_(n-2) + 2
u_(n-2) = u_(n-3) + 2
...
u_3 = u_2 + 2
u_2 = u_1 + 2
u_1 = 0
Then you just substitute back up the chain to get
u_2 = 0 + 2 = 2
u_3 = 2 + 2 = 2*2
u_4 = 2*2 + 2 = 3*2
...
u_n = (n-1)*2
Thank you so much! this really helped.
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