11⋅2⋅3+32⋅3⋅4+53⋅4⋅5+74⋅5⋅6+...
=n∑k=12k−1k⋅(k+1)⋅(k+2) =n∑k=1−12⋅k+n∑k=13k+1−n∑k=152⋅(k+2)
I do not know how to get a telescoping series from here to cancel terms.
Answer
HINT:
Note that we have
2k−1k(k+1)(k+2)=3k+1−5/2k+2−1/2k=12(1k+1−1k)+52(1k+1−1k+2)
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