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I've tried looking at video examples from my e-book, khanacademy, I can't find anything to explain this. My homeworks tutorial problems are always really confusing, they use terms when the book uses SOMETHING else. To make things worse you get 5 attempts so I'm discouraged from trying the problem again myself in case I mess up :[

Isn't a general formula the generic formula for any maclaurin series? I don't get what they even want me to do...

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  • $\begingroup$ You are doing fine so far -- they just want you to next find a simple formula for the coefficient of $x^n$ in terms of n using your answer to part 3. $\endgroup$ – user84413 Jul 16 '16 at 0:46
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you have $f(x) = 8 + 16 x + 24 x^2 + 32 x^3 + 40 x^4\cdots$ factor out an 8.

$f(x) = 8(1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4\cdots)$ look for a pattern

$f(x) = 8(1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4\cdots (n+1) x^n\cdots)\\ \sum_\limits{n=0}^\infty 8(n+1)x^n$

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  • $\begingroup$ holy crap that makes so much sense, the factoring out of the 8. thanks! $\endgroup$ – dmscs Jul 16 '16 at 0:51
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+1 for your post...I think there is a different approach to get a series representation for your function. Using geometric series: $\frac{1}{1-x}=1+x+x^2+x^3....=\sum_{n=0}^{\infty}x^n$ Now when you take the derivative and then multiply by 8, you arrive at your function. In summation form, you get: $\frac{8}{(1-x)^2}=8(1+2x+3x^2+4x^3....)=\sum_{n=1}^{\infty}8nx^{n-1}$. This summation formula is a general formula. Can you figure out the interval of convergence?

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  • $\begingroup$ They asked for the radius of convergence and I determined it was 1, using the ratio test. Thank you! $\endgroup$ – dmscs Jul 16 '16 at 0:55

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