# IYGB question help

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#1
I have worked out the answer on part (i) but am clueless in part ii
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3 years ago
#2
(Original post by Butterflyshy)
I have worked out the answer on part (i) but am clueless in part ii
Stems:

Year 1 : 1
Year 2 : 1 + 3
Year 3 : 1 + 3 + 9
Year 4 : 1 + 3 + 9 + 27

So you're looking for the nth term of this sequence. What do you notice?
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#3
(Original post by notnek)
Stems:

Year 1 : 1
Year 2 : 1 + 3
Year 3 : 1 + 3 + 9
Year 4 : 1 + 3 + 9 + 27

So you're looking for the nth term of this sequence. What do you notice?
it is geometric?
0
3 years ago
#4
(Original post by Butterflyshy)
it is geometric?
Yes, well each term is the sum of a geometric sequence.

So Year 1 is , year 2 is etc.

So the nth term of this sequence will be the same as the sum of the geometric sequence for n terms i.e. .
0
3 years ago
#5
(Original post by Butterflyshy)
it is geometric?
Yh if it was arithmetic then they wouldve had the same pattern throughout

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#6
(Original post by notnek)
Yes, well each term is the sum of a geometric sequence.

So Year 1 is , year 2 is etc.

So the nth term of this sequence will be the same as the sum of the geometric sequence for n terms i.e. .
so it is 1*(1-3^n)/1-3
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3 years ago
#7
(Original post by Butterflyshy)
so it is 1*(1-3^n)/1-3
Yes that's right although you can simplify it and write it in a nicer way.
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#8
which is then (1-3^n)/-2 which can be simplified to 1/2(3^n-1)
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#9
(Original post by notnek)
Yes that's right although you can simplify it and write it in a nicer way.
Thanks for the help I understand the question now.
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