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f(x)=(x−1)100(x−2)2(99)(x−3)3(98)⋯(x−100)100\large f(x) = (x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \cdots (x-100)^{100} f(x)=(x−1)100(x−2)2(99)(x−3)3(98)⋯(x−100)100
For f(x)f(x)f(x) as defined above, if k=f′(101)f(101)k = \dfrac {f'(101)}{f(101)}k=f(101)f′(101), find k50−97\dfrac k{50}-9750k−97.
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