Question

For the function $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+.....+\frac{x^2}{2}+x+1$
Prove that $f^{\prime }\left(1\right)=100f^{\prime }\left(0\right)$.

Answer 1

Let $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^2}{2}+x+1$
Then,
$f^{\prime }\left(x\right)=\frac{d}{dx}[\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^2}{2}+x+1]$
$=\frac{1}{100}\frac{d}{dx}{x}^{100}+\frac{1}{99}\cdot \frac{d}{dx}{x}^{99}+...+\frac{1}{2}\frac{d}{dx}{x}^2+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)$
$=\frac{1}{100}\times 100x^{99}+\frac{1}{99}+99x^{98}+...+\frac{1}{2}\times 2x+1+0$
$=x^{99}+x^{98}+...+x+1$
Then, putting $1$ and $0$ in place of $x$.
$f^{\prime }\left(1\right)=(1^{99}+1^{98}+...+1)+1$
$=1+1+1+...+99term+1$
$=99+1=100$ and $f^{\prime }\left(0\right)=1$
Hence, $f^{\prime }\left(1\right)=100$
$\because f^{\prime }\left(1\right)=100f^{\prime }\left(0\right)$ Hence Proved.