The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + . . . +^{20} C_{10}$ is

-^{20} C_{10}

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\frac{1}{2}^{20} C_{10}

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^{20} C_{10}

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Solution

The correct option is D $\frac{1}{2}^{20} C_{10}$
On putting $x = - 1$ in
$(1 + x)^{20} =^{20} C_{0} +^{20} C_{1} x + . . . + 20 C_{10} x^{10} + . . . +^{20} C_{20} x^{20}$
We get,
$0 =^{20} C_{0} -^{20} C_{1} + . . . -^{20} C_{9} +^{20} C_{10} -^{20} C_{11} + . . . +^{20} C_{20}$
$\Rightarrow 0 =^{20} C_{0} -^{20} C_{1} + . . . -^{20} C_{9} +^{20} C_{10} -^{20} C_{9} + . . . +^{20} C_{0}$ ........ $[∵^{n} C_{r} =^{n} C_{n - r}]$
$\Rightarrow 0 = 2 (^{20} C_{0} -^{20} C_{1} + . . . -^{20} C_{9}) +^{20} C_{10}$
$\Rightarrow^{20} C_{10} = 2 (^{20} C_{0} -^{20} C_{1} + . . . +^{20} C_{10})$ .... [Adding $^{20} C_{10}$ on both the sides]
$\Rightarrow^{20} C_{0} -^{20} C_{1} + . . . +^{20} C_{10} = \frac{1}{2}^{20} C_{10}$

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