## anonymous 3 years ago 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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1. anonymous

What is (a-sub-n)x^n ?

2. anonymous

any constant

3. anonymous

haven't mastered entering equations here yet haha

4. anonymous

$a_{n}$

5. anonymous

so $f(x)=\sum_{n=3}^{\infty}\frac{ 2^{n} }{ n! }\left( x-1 \right)^{n-2}$ and $g(x)=\sum_{n=1}^{\infty}\frac{ n ^{2} }{ 2^{n} }\left( x-1 \right)^{n-1}$

6. anonymous

and I need to find$f(x)+g(x)=\sum_{n=0}^{\infty}a _{n}\left( x-1 \right)^{n}$

7. anonymous

Sorry not sure try physicsforums.com

8. anonymous

ok thanks skaematik

9. anonymous

This UI is rather annoying, sorry for the way my question is stated above.

10. anonymous

$\sum_{n=1}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\left( x-1 \right)^{n}+\sum_{n=0}^{\infty}\frac{(n+1)^{2} }{ 2^{n+1} }\left( x-1 \right)^{n}$

11. anonymous

cinar, I thought something similar, but I have to force the solution to have the lower limit n=0

12. anonymous

$-1+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n$

13. anonymous

$-1+\sum_{n=0}^{\infty}a_n(x-1)^n$

14. anonymous

this should be 2 sorry $-2+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n$

15. anonymous

but it is not what you looking for right..

16. anonymous

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

17. anonymous

$-2+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }+\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n$

18. anonymous

here's what they got:

19. anonymous

20. anonymous

yeah I see now where I made a mistake..

21. anonymous

oh? btw i'm looking at your problem...no ideas for it yet tho

22. anonymous

thanks (:

23. anonymous

I though it should be related integration by part somehow but no clue yet..

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