Recurrence system

[HATLEY1997]

Any guidance on how to work this out?

 

[khansaheb]

Any guidance on how to work this out?

Please re-orient your "image" - so that we can read it easily.

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Please show us what you have tried and exactly where you are stuck.

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[blamocur]

Any guidance on how to work this out?

You could also start by making your picture readable, as suggested by @khansaheb:

 

[blamocur]

One way is to verify each of the answers against the recurrence equations.

 

[topsquark]

You could also start by making your picture readable, as suggested by @khansaheb:

View attachment 37962

[imath]a_n = \dfrac{1}{2} a_{n-1}[/imath]

[imath]a_n - \dfrac{1}{2} a_{n-1} = 0[/imath]

Now solve the associated equation
[imath]m^1 - \dfrac{1}{2}m^0 = 0[/imath]

Then [imath]a_n = a_1 m^{n-1}[/imath]

for m and use [imath]a_1 = 3[/imath].

-Dan

 

[HATLEY1997]

Thanks Dan

 

[Steven G]

blamocur gave a gave hint. I will go further with that hint.
a₂= 3/2. Only one choice gives 3/2 when n = 2. Done.