Mark the correct alternative in each of the following:

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

(a) 5050 (b) 5049 (c) 5051 (d) 50051

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Solution

$f (x) = x^{100} + x^{99} + . . . + x + 1$

Differentiating both sides with respect to x, we get

$f' (x) = \frac{d}{d x} (x^{100} + x^{99} + . . . + x + 1) = \frac{d}{d x} (x^{100}) + \frac{d}{d x} (x^{99}) + . . . + \frac{d}{d x} (x^{2}) + \frac{d}{d x} (x) + \frac{d}{d x} (1) = 100 x^{99} + 99 x^{98} + . . . + 2 x + 1 + 0 (y = x^{n} \Rightarrow \frac{d y}{d x} = n x^{n - 1}) = 100 x^{99} + 99 x^{98} + . . . + 2 x + 1$

Putting x = 1, we get

$f' (1) = 100 + 99 + 98 + . . . + 2 + 1 = \frac{100 (100 + 1)}{2} (S_{n} = \frac{n (n + 1)}{2}) = 50 \times 101 = 5050$

Hence, the correct answer is option (a).

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