A) \[{{e}^{2}}\]
B) e
C) \[{{e}^{2}}-1\]
D) \[e-1\]
E) \[{{e}^{2}}+1\]
Correct Answer: C
Solution :
\[\sum\limits_{n=1}^{\infty }{\frac{^{n}{{C}_{0}}+.......+{{\,}^{n}}{{C}_{n}}}{{{\,}^{n}}{{P}_{n}}}}\] \[=\frac{{{\,}^{1}}{{C}_{0}}+{{\,}^{1}}{{C}_{1}}}{{{\,}^{1}}{{P}_{1}}}+\frac{{{\,}^{2}}{{C}_{0}}+{{\,}^{2}}{{C}_{1}}+{{\,}^{2}}{{C}_{2}}}{{{\,}^{2}}{{P}_{2}}}+\frac{^{3}{{C}_{0}}+{{\,}^{3}}{{C}_{1}}+{{\,}^{3}}{{C}_{2}}+{{\,}^{3}}{{C}_{3}}}{{{\,}^{3}}{{P}_{3}}}\]+... \[=\frac{{{2}^{1}}}{1!}+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+.......\] \[\left( 1+\frac{2}{1!}+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+....... \right)-1\] \[={{e}^{2}}-1\].
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