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Sum of 2 power series: Given: f(x)=\sum_{n=3}^{\infty}\frac{ 2^{n} }{ n! }\left( x-1 \right)^{n-2} and g(x)=\sum_{n=1}^{\infty}\frac{ n ^{2} }{ 2^{n} }\left( x-1 \right)^{n-1} Find: f(x)+g(x)=\sum_{n=0}^{\infty}a _{n}\left( x-1 \right)^{n}

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What is (a-sub-n)x^n ?
any constant
haven't mastered entering equations here yet haha

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Other answers:

so \[f(x)=\sum_{n=3}^{\infty}\frac{ 2^{n} }{ n! }\left( x-1 \right)^{n-2}\] and \[g(x)=\sum_{n=1}^{\infty}\frac{ n ^{2} }{ 2^{n} }\left( x-1 \right)^{n-1}\]
and I need to find\[f(x)+g(x)=\sum_{n=0}^{\infty}a _{n}\left( x-1 \right)^{n}\]
Sorry not sure try
ok thanks skaematik
This UI is rather annoying, sorry for the way my question is stated above.
\[\sum_{n=1}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\left( x-1 \right)^{n}+\sum_{n=0}^{\infty}\frac{(n+1)^{2} }{ 2^{n+1} }\left( x-1 \right)^{n}\]
cinar, I thought something similar, but I have to force the solution to have the lower limit n=0
\[-1+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n\]
this should be 2 sorry \[-2+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n\]
but it is not what you looking for right..
thanks for your replies cinar, i appreciate the help. i suppose my real problem at this point is understanding how the solution from the book was reached...i'm trying to upload it now
\[-2+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }+\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n\]
here's what they got:
1 Attachment
glad to hear that..
yeah I see now where I made a mistake..
oh? btw i'm looking at your ideas for it yet tho
thanks (:
I though it should be related integration by part somehow but no clue yet..

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