\[{{\left[ 1+\frac{1}{2\,!}+\frac{1}{4\,!}+....\infty \right]}^{2}}-{{\left[ 1+\frac{1}{3\,!}+\frac{1}{5\,!}+.....\infty \right]}^{2}}=\]

\[{{\left[ 1+\frac{1}{2\,!}+\frac{1}{4\,!}+....\infty \right]}^{2}}-

1.	\[{{\left[ 1+\frac{1}{2\,!}+\frac{1}{4\,!}+....\infty \right]}^{2}}-{{\left[ 1+\frac{1}{3\,!}+\frac{1}{5\,!}+.....\infty \right]}^{2}}=\]
A.	0
B.	1
C.	\[-1\]
D.	2
Answer» C. \[-1\]

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\[{{\left[ 1+\frac{1}{2\,!}+\frac{1}{4\,!}+....\infty \right]}^{2}}-{{\left[ 1+\frac{1}{3\,!}+\frac{1}{5\,!}+.....\infty \right]}^{2}}=\]

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