Arrangment as is follows a's are there as place holders to keep the diagram intact while posting, all capacitors have = capacitance C

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The question just asks for the equvilant capacitance.

The professor hinted at the solution by writing 1/ceq = 1/c + 1/ceq where ceq is the equivelent capacitance and the answeris given as (c/2)((5)^(1/2)-1)

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Suppose you remove the two leftmost capacitors from the diagram.
What you have left is the same infinite network of capacitors, wich should
have the same capacitance.

ok I understand that, so that would leave you with 1/ceq=(2/c + 1/ceq)? because the 2 "removed" capacitors are in series

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ok I understand that, so that would leave you with 1/ceq=(2/c + 1/ceq)? because the 2 "removed" capacitors are in series

They are not in series. draw a diagram with just the first 2 capacitors and the rest of the network as Ceq

so 1/ceq=1/c + 1/ceq is what i got, which is insolvable, circuts are definatly not my strong point i guess
____||_________
aaaaaaaa|aaaaa|
aaaaaaaa=aaaaa= ceq
aaaaaaaa|aaaaa|
______________|

so 1/ceq=1/c + 1/ceq is what i got, which is insolvable, circuts are definatly not my strong point i guess
____||_________
aaaaaaaa|aaaaa|
aaaaaaaa=aaaaa= ceq
aaaaaaaa|aaaaa|
______________|

The capacity of the entire circuit Ceq is equal to the capacity of the circuit you've drawn
here, wich is one capacitor of value Ceq in parallel with a capacitor of value C, and the
equivalent capacitor to that in series with another capacitor with capacitance C

Just do like willem2 said, and remember: capacitors in series combine like resistors in parallel, and vice versa. I tried out the problem myself, and the answer you have quoted is correct: i.e., Ceq = (c/2)(5^0.5 - 1).