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Q.
If $f(x)=x^{n}+4$ then the value of $f (1)+\frac{ f ^{\prime}(1)}{1 !}+\frac{ f ^{\prime \prime}(1)}{2 !}+\ldots+\frac{ f ^{ n }(1)}{ n !}$ is
Continuity and Differentiability
Solution:
$f(1)=5, f^{\prime}(x)=n x^{n-1}$, so $f^{\prime}(1)=n$
$f^{\prime \prime}(1)=n(n-1) \ldots$ then $f^{n}(1)=1.2 \ldots . n$
$f (1) +\frac{ f ^{\prime}(1)}{1 !}+\ldots+\frac{ f ^{ n }(1)}{ n !}$
$=5+\frac{ n }{1}+\frac{ n ( n -1)}{2 !}+\ldots+\frac{ n !}{ n !} $
$=(1+1)^{ n }+4=2^{ n }+4$