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Question
If `f(x) = x^100 + x^99 .... + x + 1`, then f'(1) is equal to ______.
Options
5050
5049
5051
50051
Solution
If `f(x) = x^100 + x^99 .... + x + 1`, then f'(1) is equal to 5050.
Explanation:
Given, `f(x) = x^100 + x^99 + ... + x + 1`
∴ f'(x) = `100x^99 + 99 x^98 + ... + 1`
So, f"(1) = 100 + 99 + 98 + ... + 1
= `100/2 [2 xx 100 + (100 - 1)(-1)]`
= `50[200 - 99]`
= `50 xx 101`
= 2050
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