If r=1nTr= (2n-1)(2n+1)(2n+3)(2n+5)64, then n r=1n ( 1Tr ) is equal to :
(1) 23
Tn=Sn-Sn-1= (2n-1)(2n+1)(2n+3)8
1Tr=2 [ 1(2r-1)(2r+1)- 1(2r+1)(2r+3) ] (telescopes)
r=1n 1Tr=2 [ 13- 1(2n+1)(2n+3) ]
As n : 23
Answer: (1)
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