# A-level mathematics | sequence |

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#1
Anyone can help me with this problem.Consider the sequence 1,1,3,(1/3),9,(1/9),27,(1/27),81,(1/81),.......Show that the sum of the first 2n terms of the sequence is 1/2[2-(3^n)-(3^(n-1)].
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11 months ago
#2
(Original post by Lami13)
Anyone can help me with this problem.Consider the sequence 1,1,3,(1/3),9,(1/9),27,(1/27),81,(1/81),.......Show that the sum of the first 2n terms of the sequence is 1/2[2-(3^n)-(3^(n-1)].
What do you notice about the sequence(s)?

The hint is in my question.
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#3
(Original post by mqb2766)
What do you notice about the sequence(s)?

The hint is in my question.
I still don’t know how to prove it.
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11 months ago
#4
(Original post by Lami13)
I still don’t know how to prove it.
What do you notice about the sequence(s)?

The aim is to help you do the question, not do it for you.
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11 months ago
#5
Split the 2n sequence into two sequences of length n
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#6
(Original post by Matureb)
Split the 2n sequence into two sequences of length n
Hahaha got it. Thanks 😊
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#7
Anyone may help me solve these 2 questions
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11 months ago
#8
(Original post by Lami13)
Anyone may help me solve these 2 questions
Looks like you need to represent the circles as a sequence (geometric progression?). Try working out the intiial value and the ratio for the radius (circumference) / area? I suppose looking at the side length : radius for the initial triangle would be good, then thinking about how the triangle gets divided up at each stage?

The snowflake is well described online. What have you done / what problems are you having?
Last edited by mqb2766; 11 months ago
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