If $x^{100} + 2 x^{99} + k$ is divisible by $(x + 1)$ , then the value of $k$ is

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Solution

The correct option is A
1

Since $(x + 1)$ is a factor, $x = - 1$ will make the given expression zero.

$∴ (- 1)^{100} + 2 \times (- 1)^{99} + k = 0 \Rightarrow 1 - 2 + k = 0 \Rightarrow k = 1$

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