Question

If $f\left(x\right)=1+x+\frac{x^2}{2}+....+\frac{x^{200}}{200}$
, then f '(1) is equal to

• A. $\frac{1}{200}$
• B. 200
• C. 100
• D. 0.0

Answer 1

The correct option is B 200

Given that $f\left(x\right)=1+x+\frac{x^2}{2}+....+\frac{x^{200}}{200}\therefore f\text{'}\left(x\right)=\frac{d}{\mathrm{dx}}\left(1+x+\frac{x^2}{2}+....+\frac{x^{200}}{200}\right\}\Rightarrow f\text{'}\left(x\right)=0+1+\frac{2x}{2}+....+\frac{200x^{199}}{200}\therefore f\text{'}\left(1\right)=0+1+1+....+1\left(200\mathrm{times}\right)\Rightarrow f\text{'}\left(1\right)=200$