Sum the first n terms of the series: 1⋅3⋅22+2⋅4⋅32+3⋅5⋅42+⋯.
This was asked under the heading using method of difference and the answer given was Sn=110n(n+1)(n+2)(n+3)(2n+3).
First, I get Un=n(n+2)(n+1)2.
Then I tried to make Un=Vn−Vn−1 in order to get Sn=Vn−V0. But I really don't know how can I figure this out.
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