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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? \[\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^{\infty} a_n\]

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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?
I'm trying to start at n=0
\[\sum_{n=1}^{\infty}a_nnx^n \rightarrow \sum_{n=0}^{\infty}a_nnx^n\]

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Other answers:

I looked at example 1 a. but I'm not trying to do an index shift :C
Well do you need to start at 0? because if you want n = 0 you can just make it at the bottom n = 0
why though?
why is that all I need to do?
\[\sum_{n =0}^{\infty}\]
because in your equation it would work for not getting an answer anyway
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.
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.
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?
it's right the left part
but if you made that n = 0
on the left and n= 1 on the right
it would be \[\alpha _{0}*0*x^0\]
OMG! I think something just clicked!
Yeah, you see where im going with it
no wait, just a second
You can have it start at 0.. but it doesnt change anything if you start at 1 in this case
it's like putting 0 + \[\sum_{n=1}^{\infty} (insert rest here)\]
for n = 0
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.
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?
I'm honestly trying hard to get this...I think we're almost there...sorry =/
\[\sum_{n=0}^{\infty}a _{n}nx ^{n} = 0 + \sum_{n=1}^{\infty}a _{n}nx ^{n}\]
using the rule that 0 multiplied by anything is 0
Thank you, finally!
Lol it's ok, glad i could help

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