- #1

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Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

my A=2

my r= -1/4

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- Thread starter mattmannmf
- Start date

- #1

- 172

- 0

Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

my A=2

my r= -1/4

- #2

Mark44

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[tex]\sum_{n = 1}^{\infty} 2 (-1/4)^{n - 1}[/tex]

Write a few terms of this series and see if you can adjust the starting index.

- #3

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i have no idea what you mean by that

- #4

Hurkyl

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Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

my A=2

my r= -1/4

You have

2(-1/4)^{n-1}

and you wantA r^{n-1}

?Why not set them equal and try solving for A and r?

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- #6

Hurkyl

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- #7

Char. Limit

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Well, just change n-1 into n using exponent rules.

- #8

Mark44

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