The sum of the series C0 20 – C1 20 +C2 20 – C3 20 + … – … +C10 20 is

We have (1+x)20=C0 20+C1 20x+C2 20nx2+…+C20 20x20 Put x=-1⇒0=C0 20-C1 20+C2 20-C3 20+C4 20-….-C9 20+C10 20-C11 20+C13 20+…-C19 20+C20 20⇒0=2C0 20-C1 20+C2 20-C3 20+…-C9 20+C10 20-C10 20 ∵Cr=Cn-r n n⇒12C10 20=C0 20-C1 20+C2 20-C3 20+…+C10 20

The sum of the series C0 20 – C1 20 +C2 20 – C3 20 + … – … +C10 20 is

We have (1+x)20=C0 20+C1 20x+C2 20nx2+…+C20 20x20 Put x=-1⇒0=C0 20-C1 20+C2 20-C3 20+C4 20-….-C9 20+C10 20-C11 20+C13 20+…-C19 20+C20 20⇒0=2C0 20-C1 20+C2 20-C3 20+…-C9 20+C10 20-C10 20 ∵Cr=Cn-r n n⇒12C10 20=C0 20-C1 20+C2 20-C3 20+…+C10 20

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

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$- {}^{20}C_{10}$

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

${}^{20}C_{10}$

answer is B.

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Detailed Solution

$We have (1 + x)^{20} = {}^{20}C_{0} + {}^{20}C_{1} x + {}^{20 n}C_{2} x^{2} + \dots + {}^{20}C_{20} x^{20} Put x = - 1$

$\Rightarrow 0 = {}^{20}C_{0} - {}^{20}C_{1} + {}^{20}C_{2} - {}^{20}C_{3} + {}^{20}C_{4} - \dots . - {}^{20}C_{9} + {}^{20}C_{10} - {}^{20}C_{11} + {}^{20}C_{13} + \dots - {}^{20}C_{19} +$ ${}^{20}C_{20}$

$\begin{array}{l} \Rightarrow 0 = 2 [{}^{20}C_{0} - {}^{20}C_{1} + {}^{20}C_{2} - {}^{20}C_{3} + \dots - {}^{20}C_{9} + {}^{20}C_{10}] - {}^{20}C_{10} (∵ {}^{n}{C_{r} = {}^{n}{C_{n - r}}}) \\ \Rightarrow \frac{1}{2} {}^{20}C_{10} = {}^{20}C_{0} - {}^{20}C_{1} + {}^{20}C_{2} - {}^{20}C_{3} + \dots + {}^{20}C_{10} \end{array}$