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

#### 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).