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Q.

If Tn=(n2+1)n ! and Sn=T1+T2++Tn, let T50S50=ab, where a and   b are relatively prime natural numbers, then value of ba  is equal to

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a

48

b

50

c

39

d

49

answer is A.

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Detailed Solution

A We have   

Tr=[(r+1)(r+1)!r.r!][(r+1)!r!]

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