JEE maths#6

Calculus Level 3

f(x)=(x1)100(x2)2(99)(x3)3(98)(x100)100\large f(x) = (x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \cdots (x-100)^{100}

For f(x)f(x) as defined above, if k=f(101)f(101)k = \dfrac {f'(101)}{f(101)}, find k5097\dfrac k{50}-97.


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