Basic maths-calculus-Q14

+1 vote

If f(x) = x^{100} + x^{99} +...+x + 1, then f'(1) is equal to

(A). 5050

(B). 5049

(C). 5051

(D). 50051

asked May 19 in Basic Maths by gbeditor (44,500 points)
reshown May 24 by gbeditor

1 Answer

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Best answer

By summation of geometric progression,

1+x+x^2 + \dots + x^n = 1(x^{n+1} - 1)/(x-1)

1+x+x^2 + \dots + x^{100} = (x^{101} - 1)/(x-1)

f(x) = (x^{101} - 1)/(x-1)

f'(x) = \frac{[ (x-1)(101x^{100} ) - (x^{101} - 1) ]}{(x-1)^2}

Put x =1,

f'(1) = 5050

answered May 27 by deepak-gatebook (112,390 points)
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