A) \[{{(2019)}^{n}}\]
B) \[{{(2020)}^{n}}\]
C) \[{{(2020)}^{n}}-1\]
D) \[n\,{{(2019)}^{n}}\]
Correct Answer: C
Solution :
[c]\[f(x)={{(x+2019)}^{n}}\] |
\[f(0)={{(2019)}^{n}}\] |
\[f'(x)=n{{(x+2019)}^{n-1}},\] |
\[f'(0)=n{{(2019)}^{n-1}}\] |
\[f''(x)=n(x-1){{(x+2019)}^{n-2}},\]\[f''(0)=n(n-1){{(2019)}^{x-2}}\] |
\[{{f}^{n-1}}(n)=n\,(n-1)...1\] |
\[{{f}^{n-1}}(0)=n(n-1)\times ...1\] |
\[f(0)+f'(0)+\frac{f''0}{2!}+...+\frac{{{f}^{n-1}}(0)}{(n-1)!}\] |
\[{{(2019)}^{n}}+{}^{n}{{C}_{1}}{{(2019)}^{n-1}}+{}^{n}{{C}_{2}}{{(2019)}^{n-2}}\]\[+...+{}^{n}{{C}_{n-1}}2019{{+}^{n}}{{C}_{n-1}}\] |
\[={{(2019+1)}^{n}}-1\]\[={{(2020)}^{n}}-1\] |
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