Wednesday, April 27, 2016

real analysis - An alternative way to find the sum of this series?

420+4.720.30+4.7.1020.30.40+...



Now I have tried to solve this in a usual way, first find the nth term tn.



tn= 110(1+32) + 1102(1+32)(1+63) + ...+ 110n(1+32)(1+63)...(1+3nn+1)



=110n(1+2rr+1) , r=1,2,..,n




=(31015(r+1))
thus, tn= (x-a2)(x-a3)...(x-an+1), x=310, a=15



Now to calculate Sn, I have to find the product tn, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?

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