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Mathematics Question on Binomial theorem

The sum of the series 20C0−20C1+20C2−20C3+.....−....+20C10{^{20}C_0} - {^{20}C_1} + {^{20}C_2} - {^{20}C_3} + ..... - .... + {^{20}C_{10}} is

A

0

B

20C10{^{20}C_{10}}

C

−20C10{^{ - 20}C_{10}}

D

1220C10 \frac{1}{2} {^{20}C_{10}}

Answer

1220C10 \frac{1}{2} {^{20}C_{10}}

Explanation

Solution

(1+x)20=20C0+20C1x+...+20C10x10+...+20C20x20\left(1+x\right)^{20}=^{20}C_{0}+^{20}C_{1}x+...+^{20}C_{10}x^{10}+...+^{20}C_{20}x^{20} put x=−1,x = - 1, 0=20C0−20C1+...−20C9+20C11+...+20C200=^{20}C_{0}-^{20}C_{1}+...-^{20}C_{9}+^{20}C_{11}+...+^{20}C_{20} 0=2(20C0−20C1+...−20C9)=20C100=2\left(^{20}C_{0}-^{20}C_{1}+...-^{20}C_{9}\right)=^{20}C_{10} ⇒20C0−20C1+...+20C10=1220C10.\Rightarrow ^{20}C_{0}-^{20}C_{1}+...+^{20}C_{10}=\frac{1}{2}^{20}C_{10}.