Question

If $f\left(x\right)={(1+x)}^8$ then $f\left(0\right)+\frac{f^{\prime }\left(0\right)}{1!}+\frac{f^{\prime \prime }\left(0\right)}{2!}+....+\frac{f^{VIII}\left(0\right)}{8!}$ is equal to

• A. $256$
• B. $128$
• C. $1+1!+\frac{1}{2!}+....+\frac{1}{8!}$
• D. an irrational number

Answer 1

The correct option is A $256$

Taylor series for $f\left(x\right)$ at $x=0$ is
$(1+x{)}^8=f(0)+\frac{f^{\prime }\left(0\right)}{1!}(x-0)+\frac{f^{\prime \prime }\left(0\right)}{2!}(x-0{)}^2+...........\frac{f^{viii}\left(0\right)}{8!}(x-0{)}^8$
Now substitute $x=1$
$f\left(0\right)+\frac{f^{\prime }\left(0\right)}{1!}+\frac{f^{\prime \prime }\left(0\right)}{2!}+...........\frac{f^{viii}\left(0\right)}{8!}=(2{)}^8=256$