Question

Find the derivative of $f\left(x\right)=1+x+x^2+--+x^{50}atx=1$.

Answer 1

$f\left(x\right)=1+x+x^2+.......+x^{50}$

For $f^{\prime }\left(x\right)$ at $x=1$

$f^{\prime }\left(x\right)=0+1+2x+3x^2+.....+50x^{49}$

At $x=1$,

$f^{\prime }\left(1\right)=1+2\left(1\right)+3{\left(1\right)}^2+.....+50{\left(1\right)}^{49}$

$\Rightarrow f^{\prime }\left(1\right)=1+2+3+.....+50$

As we know that sum of $n$ natural numbers is given as-

$S=\frac{n(n+1)}{2}$

$\therefore f^{\prime }\left(1\right)=\frac{50(50+1)}{2}=1275$

Hence the value of $f^{\prime }\left(x\right)$ at $x=1$ is $1275$.