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Question
Mark the correct alternative in of the following:
If \[f\left( x \right) = x^{100} + x^{99} + . . . + x + 1\] then \[f'\left( 1 \right)\] is equal to
Options
5050
5049
5051
50051
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Solution
\[f\left( x \right) = x^{100} + x^{99} + . . . + x + 1\]
Differentiating both sides with respect to x, we get \[f'\left( x \right) = \frac{d}{dx}\left( x^{100} + x^{99} + . . . + x + 1 \right)\]
\[ = \frac{d}{dx}\left( x^{100} \right) + \frac{d}{dx}\left( x^{99} \right) + . . . + \frac{d}{dx}\left( x^2 \right) + \frac{d}{dx}\left( x \right) + \frac{d}{dx}\left( 1 \right)\]
\[ = 100 x^{99} + 99 x^{98} + . . . + 2x + 1 + 0 \left( y = x^n \Rightarrow \frac{dy}{dx} = n x^{n - 1} \right)\]
\[ = 100 x^{99} + 99 x^{98} + . . . + 2x + 1\]
Putting x = 1, we get
\[f'\left( 1 \right) = 100 + 99 + 98 + . . . + 2 + 1\]
\[ = \frac{100\left( 100 + 1 \right)}{2} \left( S_n = \frac{n\left( n + 1 \right)}{2} \right)\]
\[ = 50 \times 101\]
\[ = 5050\]
Hence, the correct answer is option (a).
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