If f x -1100-100-x99-99-x98-98-x2-2- x -1- then f -1-A- 100B- 10C- 1D- 0

The correct option is A 100 f’(x) = x^99 + x^98 + x^97 + ⋯ + 1 ∴ f'(1) = 100 ...

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Integration to Solve Modified Sum of Binomial Coefficients

If f x =1100/...

Question

If $f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \frac{x^{98}}{98} + \dots + \frac{x^{2}}{2} + x + 1$ , then $f^{'} (1) =$

100

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Solution

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Similar questions

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \frac{x^{98}}{98}

, s h o w t h a t

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . . . . + \frac{x^{2}}{2} + x + 1

f (x)

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . . . + \frac{x^{2}}{2} + x + 1

f^{'} (0) =

If the function f(x) defined by $f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^{2}}{2} + x + 1, t h e n f^{'} (0) =$

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . . . . . . . . . + \frac{x^{2}}{2} + x + 1

f^{'} (1) =

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