Find S=17+1⋅37⋅9+1⋅3⋅57⋅9⋅11+⋯ upto 20 terms
I first multiplied and divided S with 1⋅3⋅5 S15=11⋅3⋅5⋅7+1⋅31⋅3⋅5⋅7⋅9+1⋅3⋅51⋅3⋅5⋅7⋅9⋅11+⋯ Using the expansion of (2n)! 1⋅3⋅5⋯(2n−1)=(2n)!2nn! S=15[20∑r=1(2r)!2rr!(2(r+3))!2r+3(r+3)!] S=15⋅8⋅[20∑r=1(2r)!r!⋅(r+3)!(2r+6)!] S=1520∑r=11(2r+5)(2r+3)(2r+1)
How can I solve the above expression? Or is there an simpler/faster method?
Answer
Hint:
1(2r+5)(2r+3)(2r+1)=14(1(2r+3)(2r+1)−1(2r+5)(2r+3))
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