Question

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

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

Answer 1

The correct option is A $5050$

Given: $f\left(x\right)=x^{100}+x^{99}+....+x+1$

To Find: $f^{{}^{\prime }}\left(1\right)$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=100x^{99}+99x^{99}+.....+2x+1$
$(\frac{d{x}^n}{dx}=n{x}^{n-1})$

put $x=1$, we get

$\Rightarrow f^{{}^{\prime }}\left(1\right)=100+99+.....+2+1$
$\therefore f^{{}^{\prime }}\left(1\right)=\frac{100}{2}[2\times 100+(100-1\left)\right(-1\left)\right]$

$\Rightarrow f^{{}^{\prime }}\left(1\right)=50\times 101$
$\Rightarrow f^{{}^{\prime }}\left(1\right)=5050$
Hence, the correct option is (a)