- #1

mattmannmf

- 172

- 0

E (-1)^(n-1) * (n/(2^{n-1}))

By using the power series 1/ (1+x) = E (-1)^n *x^n,

I am unsure of what to do...

Do I take the derivative with respect to x of both sides?

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- Thread starter mattmannmf
- Start date

- #1

mattmannmf

- 172

- 0

E (-1)^(n-1) * (n/(2^{n-1}))

By using the power series 1/ (1+x) = E (-1)^n *x^n,

I am unsure of what to do...

Do I take the derivative with respect to x of both sides?

- #2

mattmannmf

- 172

- 0

1/ (1+x) = E (-1)^n *x^n

now i took the derivative with respect to x:

-(1+x)^-2 = E (-1)^n * n*x^(n-1)

Then i substituted x by 1/2 getting:

-(1+(1/2))^-2 = E (-1)^(n-1) * (n/ 2^(n-1))

Does my math look correct? now my sum should be pretty much what i get from this equation:

-(1+x)^-2

but with the substituted 1/2 for x.. right?

- #3

Cyosis

Homework Helper

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You can write your original sum as:

[tex]-2 \sum_n (-1)^n n \left(\frac{x}{2}\right)^n[/tex]

Compare this to the given power series.

- #4

Cyosis

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- #5

mattmannmf

- 172

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(1+(1/2))^-2 = E (-1)^(n-1) * (n/ 2^(n-1))

and to find the sum would be to just do the math for:

(1+(1/2))^-2 = .444

- #6

Cyosis

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Yes although I would write 4/9 instead.

- #7

mattmannmf

- 172

- 0

thank you so much!

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