• # question_answer If $f(x)={{(x+2019)}^{n}},$ where x is a real variable and n is a positive integer, then the value of $f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f'''(0)}{3!}+...+\frac{{{f}^{n-1}}(0)}{(n-1)!}$ A) ${{(2019)}^{n}}$                    B) ${{(2020)}^{n}}$ C) ${{(2020)}^{n}}-1$     D) $n\,{{(2019)}^{n}}$

[c]  $f(x)={{(x+2019)}^{n}}$ $f(0)={{(2019)}^{n}}$ $f'(x)=n{{(x+2019)}^{n-1}},$ $f'(0)=n{{(2019)}^{n-1}}$ $f''(x)=n(x-1){{(x+2019)}^{n-2}},$$f''(0)=n(n-1){{(2019)}^{x-2}}$
 ${{f}^{n-1}}(n)=n\,(n-1)...1$ ${{f}^{n-1}}(0)=n(n-1)\times ...1$ $f(0)+f'(0)+\frac{f''0}{2!}+...+\frac{{{f}^{n-1}}(0)}{(n-1)!}$ ${{(2019)}^{n}}+{}^{n}{{C}_{1}}{{(2019)}^{n-1}}+{}^{n}{{C}_{2}}{{(2019)}^{n-2}}$$+...+{}^{n}{{C}_{n-1}}2019{{+}^{n}}{{C}_{n-1}}$ $={{(2019+1)}^{n}}-1$$={{(2020)}^{n}}-1$