Jrucks
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The first part of the question asks to derive the first five terms of the series expansion for ex. Next, find an expression for the nth term of the infinite series: (1/1!) + (4/2!) + (7/3!) + (10/4!) + … Which is (3n - 2)(n!). It then asks to sum the infinite series, which is what I am stuck on. I know you have to relate It to the series expansion for e^x.
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mqb2766
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(Original post by Jrucks)
The first part of the question asks to derive the first five terms of the series expansion for ex. Next, find an expression for the nth term of the infinite series: (1/1!) + (4/2!) + (7/3!) + (10/4!) + ...
So what are you stuck with?
a) Are you doing a power series (no differentiation) or a taylor series (differentiation)?
b) The nth term isn't hard. The denominator should be straightforward and if you difference the numerators, you should be able to spot what type of sequence the individual terms are.
Post your attempt and describe where you're stuck?
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Jrucks
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(Original post by mqb2766)
So what are you stuck with?
a) Are you doing a power series (no differentiation) or a taylor series (differentiation)?
b) The nth term isn't hard. The denominator should be straightforward and if you difference the numerators, you should be able to spot what type of sequence the individual terms are.
Post your attempt and describe where you're stuck?
I accidentally pressed the enter button midway through typing this; edited now. It is a Maclaurin expansion.
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RDKGames
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(Original post by Jrucks)
The first part of the question asks to derive the first five terms of the series expansion for ex. Next, find an expression for the nth term of the infinite series: (1/1!) + (4/2!) + (7/3!) + (10/4!) + … Which is (3n - 2)(n!). It then asks to sum the infinite series, which is what I am stuck on. I know you have to relate It to the series expansion for e^x.
\displaystyle e^x = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

\displaystyle \sum_{n=0}^{\infty} \dfrac{3n - 2}{n!} = -2 + \sum_{n=1}^{\infty} \dfrac{3n - 2}{n!} = -2 + 3\sum_{n=1}^{\infty} \dfrac{1}{(n-1)!} - 2 \sum_{n=1}^{\infty} \dfrac{1}{n!}


Do they look similar to the series for e^x? Try relating them to it.
Last edited by RDKGames; 5 days ago
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mqb2766
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(Original post by Jrucks)
I accidentally pressed the enter button midway through typing this; edited now. It is a Maclaurin expansion.

RDKGames got there before me :-)
Last edited by mqb2766; 5 days ago
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Jrucks
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(Original post by RDKGames)
\displaystyle e^x = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

\displaystyle \sum_{n=0}^{\infty} \dfrac{3n - 2}{n!} = -2 + \sum_{n=1}^{\infty} \dfrac{3n - 2}{n!} = -2 + 3\sum_{n=1}^{\infty} \dfrac{1}{(n-1)!} - 2 \sum_{n=1}^{\infty} \dfrac{1}{n!}


Do they look similar to the series for e^x? Try relating them to it.
Got it now; it was not spotting that (n)/(n!) can be written as (1)/(n-1)! that hindered my progress. Thanks!
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