Question

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

• A. 5050
• B. 5049
• C. 5051
• D. 50051

Answer 1

The correct option is A 5050

$f\left(x\right)=x^{100}+x^{99}+...+x+1$
Differentiating both sides with respect to x, we get
$f^{\prime }\left(x\right)=\frac{d}{dx}(x^{100}+x^{99}+...+x+1)$
$f^{\prime }\left(x\right)=\frac{d}{dx}\left(x^{100}\right)+\frac{d}{dx}\left(x^{99}\right)+...+\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)$
$=100x^{99}+99x^{98}+...+2x+1+0$ $(y=x^n\Rightarrow \frac{dy}{dx}=n{x}^{n-1})$
$=100x^{99}+99x^{98}+...+2x+1$
Putting x=1, we get
$f^{\prime }\left(1\right)=100+99+98+...+2+1$
$=\frac{100(100+1)}{2}$ $S_n=\frac{n(n+1)}{2}$
$=50\times 101=5050$
Hence, the correct answer is option (a).