I need to find the sum of n terms of the series
11⋅3+21⋅3⋅5+31⋅3⋅5⋅7+⋯
And I've no idea how to move on. It doesn't look like an arithmetic progression or a geometric progression. As far as I can tell it's not telescoping. What do I do?
Answer
It is telescoping. Consider that:
11⋅3=12(11−13),21⋅3⋅5=12(11⋅3−13⋅5),
31⋅3⋅5⋅7=12(11⋅3⋅5−13⋅5⋅7),…
so:
n∑k=1k(2k+1)!!=12(1−1(2n+1)!!).
As usual, (2k+1)!! stands for 1⋅3⋅5⋅…⋅(2k+1).
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