Oh I think I get the idea of stripping out a term now.
Simply by taking out an \[a_1\] changes where the sum starts? Yes?
\[\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^{\infty} a_n\]

- anonymous

- jamiebookeater

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- anonymous

so when I have \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=2}^{\infty}a_nnx^n\]
but it's not n=0? Why?

- anonymous

I'm trying to start at n=0

- anonymous

\[\sum_{n=1}^{\infty}a_nnx^n \rightarrow \sum_{n=0}^{\infty}a_nnx^n\]

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## More answers

- anonymous

I looked at example 1 a.
http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx
but I'm not trying to do an index shift :C

- anonymous

Well do you need to start at 0? because if you want n = 0 you can just make it at the bottom n = 0

- anonymous

why though?

- anonymous

why is that all I need to do?

- anonymous

\[\sum_{n =0}^{\infty}\]

- anonymous

because in your equation it would work for not getting an answer anyway

- anonymous

because multiplying by 0 just leaves the term out basically, so you can have it as n=0 but you won't get an actual term out.

- anonymous

if you are trying to get a term out there have to be different conditions met in your equation otherwise it won't work that way. But it looks fine for getting a term out if it's not n = 0.

- anonymous

by that do you mean that
\[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\]
It leaves the \[a_1\]out?

- anonymous

no

- anonymous

:(

- anonymous

it's right the left part

- anonymous

but if you made that n = 0

- anonymous

on the left and n= 1 on the right

- anonymous

it would be \[\alpha _{0}*0*x^0\]

- anonymous

OMG! I think something just clicked!

- anonymous

Yeah, you see where im going with it

- anonymous

no wait, just a second

- anonymous

You can have it start at 0.. but it doesnt change anything if you start at 1 in this case

- anonymous

\[\sum_{n=0}^{\infty}a_nnx^n=0\]

- anonymous

it's like putting 0 + \[\sum_{n=1}^{\infty} (insert rest here)\]

- anonymous

for n = 0

- anonymous

I guess if you want to start at 0, it has to be fixed at 0 and you can't change that value for it so it's usually left out of summations and n starts at 1. There are situations where you can have n start at 0 but in you equation those conditions for having a different term at n = 0 is not met.

- anonymous

I'm a very visual learner.
Where you said: "on the left and n= 1 on the right"
\[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\]
^^^Is wrong I know, but would you mind writing what you mean?

- anonymous

I'm honestly trying hard to get this...I think we're almost there...sorry =/

- anonymous

\[\sum_{n=0}^{\infty}a _{n}nx ^{n} = 0 + \sum_{n=1}^{\infty}a _{n}nx ^{n}\]

- anonymous

using the rule that 0 multiplied by anything is 0

- anonymous

Thank you, finally!

- anonymous

Lol it's ok, glad i could help

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