11!.100!+13!.98!+15!.96!+......+1101!=

1(101)![101C1+101C3+−−−−+101C101]=2100(101)!

11!.100!+13!.98!+15!.96!+......+1101!=

1(101)![101C1+101C3+−−−−+101C101]=2100(101)!

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$\frac{1}{1! .100!} + \frac{1}{3! .98!} + \frac{1}{5! .96!} + ...... + \frac{1}{(101)!} =$

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$\frac{2^{100}}{100!}$

$\frac{2^{101}}{101!}$

$\frac{2^{100}}{(101)!}$

$\frac{2^{51}}{(101)!}$

answer is C.

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$\frac{1}{(101)!} [101_{C_{1}} + 101_{C_{3}} + - - - - + 101_{C_{101}}] = \frac{2^{100}}{(101)!}$

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11!.100!+13!.98!+15!.96!+......+1101!=

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$\frac{1}{1! .100!} + \frac{1}{3! .98!} + \frac{1}{5! .96!} + ...... + \frac{1}{(101)!} =$