I'm trying to find general term of this series but when I run the equation on wolfram alpha it's given below general term. But I dont understant how can I transform the equation like that.
Can you help me for solution? Any source or reccomend.
Thx
Find General Term
- Thread starter eboolean
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I'm trying to find general term of this series but when I run the equation on wolfram alpha it's given below general term. But I dont understant how can I transform the equation like that.
Can you help me for solution? Any source or reccomend.
Thx
- Joined
- Nov 24, 2012
- Messages
- 3,021
What I would do is factorize the denominator of the summand:
[MATH]\frac{1}{x^3+3x^2-x-3}=\frac{1}{x^2(x+3)-(x+3)}=\frac{1}{(x+1)(x-1)(x+3)}[/MATH]
And then use partial fractions:
[MATH]\frac{1}{(x+1)(x-1)(x+3)}=\frac{1}{8}\left(\frac{1}{x+3}+\frac{1}{x-1}-\frac{2}{x+1}\right)[/MATH]
Now you should be able to get some telescoping to happen...
[MATH]\frac{1}{x^3+3x^2-x-3}=\frac{1}{x^2(x+3)-(x+3)}=\frac{1}{(x+1)(x-1)(x+3)}[/MATH]
And then use partial fractions:
[MATH]\frac{1}{(x+1)(x-1)(x+3)}=\frac{1}{8}\left(\frac{1}{x+3}+\frac{1}{x-1}-\frac{2}{x+1}\right)[/MATH]
Now you should be able to get some telescoping to happen...