Question

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

Answer 1

The given function is
$f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdot \cdot \cdot +\frac{x^2}{2}+x+1$
$\Rightarrow f^{{}^{\prime }}\left(x\right)=x^{99}+x^{98}+\cdot \cdot \cdot +x+1$
$\Rightarrow f^{{}^{\prime }}\left(1\right)=1^{99}+1^{98}+\cdot \cdot \cdot +1+1={[1+1+\cdot \cdot \cdot +1+1]}_{100\text{terms}}=1\times 100=100$
and $f^{\prime }\left(0\right)=1$
Thus $f^{{}^{\prime }}\left(1\right)=100\times f^1\left(0\right)$