The sum of the series 20C0− 20C1+ 20C2− 20C3+⋯+ 20C10 is

(1−x)20= 20C0− 20C1x+ 20C2x2− 20C3x3+⋯+ 20C20x20 put x=1 20C0− 20C1+ 20C2− 20C3+⋯+ 20C20=0 20C0− 20C1+ 20C2− 20C3+⋯+ 20C10= 20C11− 20C12+ 20C13+ 20C14⋯− 20C20 = 20C9− 20C8+ 20C7⋯ 20C0 = 20C10− 20C10+ 20C9− 20C8⋯− 20C0 2[ 20C0− 20C1+ 20C2−⋯+ 20C10]= 20C10 ∴ 20C0− 20C1+ 20C2− 20C3.....+ 20C10=12 20C10

The sum of the series 20C0− 20C1+ 20C2− 20C3+⋯+ 20C10 is

(1−x)20= 20C0− 20C1x+ 20C2x2− 20C3x3+⋯+ 20C20x20 put x=1 20C0− 20C1+ 20C2− 20C3+⋯+ 20C20=0 20C0− 20C1+ 20C2− 20C3+⋯+ 20C10= 20C11− 20C12+ 20C13+ 20C14⋯− 20C20 = 20C9− 20C8+ 20C7⋯ 20C0 = 20C10− 20C10+ 20C9− 20C8⋯− 20C0 2[ 20C0− 20C1+ 20C2−⋯+ 20C10]= 20C10 ∴ 20C0− 20C1+ 20C2− 20C3.....+ 20C10=12 20C10

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

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^{20} C_{10}

- (^{20} C_{10})

\frac{1}{2} . (^{20} C_{10})

answer is C.

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Detailed Solution

{(1 - x)}^{20} =^{20} C_{0} -^{20} C_{1} x +^{20} C_{2} x^{2} -^{20} C_{3} x^{3} + \dots +^{20} C_{20} x^{20}

$put x = 1$

$^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{20} = 0$

$^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10} =^{20} C_{11} -^{20} C_{12} +^{20} C_{13} +^{20} C_{14} \dots -^{20} C_{20}$