Sick-quences!! (6)

Algebra Level 4

Tr T_ r is a sequence such that

r=1nTr=n(n+1)(n+2)(n+3)8. \sum_{r =1 } ^ n T_r = \frac{ n(n+1)(n+2)(n+3) } { 8 }.

r=1111Tr=ab \displaystyle \sum_{r=1}^{11} \frac{1}{T_r } = \frac{a}{b}, where a,ba, b are coprime positive integers. Find a+b a + b .

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