I am not sure I understand unfortunatelyThere is a popular trick for this kind of sums: [math]x_n = \frac{1}{3n(3n+3)} = \frac{1}{3}\left(\frac{1}{3n}-\frac{1}{3n+3}\right)[/math]
This is called a "telescoping" series. Do a Google search. Let us know what did you find. Carefully, look at response#4 and think a bit.I am not sure I understand unfortunately
Watched it and managed to solve the problem. I got 1/9. Thank youuThis is called a "telescoping" series. Do a Google search. Let us know what did you find. Carefully, look at response#4 and think a bit.
reference:
Telescoping series (video) | Series | Khan Academy
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Partial fraction decomposition:Watched it and managed to solve the problem. I got 1/9. Thank youu