The value of $^{40} C_{0} +^{40} C_{1} +^{40} C_{2} + . . . . . . +^{40} C_{20}$ is?

2^{40} + \frac{40!}{(20!)^{2}}

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2^{39} - \frac{1}{2} \times \frac{40!}{(20!)^{2}}

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2^{39} +^{40} C_{20}

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None of these

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Solution

The correct option is D None of these
$^{40} C_{0} +^{40} C_{1} +^{40} C_{2} + . . . . +^{40} C_{20}$
$= \frac{1}{2} [2 \cdot^{40} C_{0} + 2 \cdot^{40} C_{1} + 2 \cdot^{40} C_{2} + . . . . . + 2 \cdot^{40} C_{20}]$
$= \frac{1}{2} [(^{40} C_{0} +^{40} C_{40}) + (^{40} C_{1} +^{40} C_{39}) + . . . . . + (^{40} C_{19} +^{40} C_{21}) + 2^{40} C_{20}]$
$= \frac{1}{2} [{^{40} C_{0} +^{40} C_{1} +^{40} C_{2} + . . . . . . +^{40} C_{19} +^{40} C_{20} +^{40} C_{21} + . . . . +^{40} C_{40}} +^{40} C_{20}]$
$= \frac{1}{2}$ $[2^{40} + \frac{40!}{(20!)^{2}}] = 2^{39} + \frac{1}{2} \frac{40!}{(20!)^{2}}$ .

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