First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

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

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

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

0

Solution

$We have (1 + x)^{20} = {}^{20}C_{0} + {}^{20}C_{1} x + {}^{20 n}C_{2} x^{2} + \dots + {}^{20}C_{20} x^{20} Put x = - 1$
$\Rightarrow 0 = {}^{20}C_{0} - {}^{20}C_{1} + {}^{20}C_{2} - {}^{20}C_{3} + {}^{20}C_{4} - \dots . - {}^{20}C_{9} + {}^{20}C_{10} - {}^{20}C_{11} + {}^{20}C_{13} + \dots - {}^{20}C_{19} +$ ${}^{20}C_{20}$
$\begin{array}{l} \Rightarrow 0 = 2 [{}^{20}C_{0} - {}^{20}C_{1} + {}^{20}C_{2} - {}^{20}C_{3} + \dots - {}^{20}C_{9} + {}^{20}C_{10}] - {}^{20}C_{10} (∵ {}^{n}{C_{r} = {}^{n}{C_{n - r}}}) \\ \Rightarrow \frac{1}{2} {}^{20}C_{10} = {}^{20}C_{0} - {}^{20}C_{1} + {}^{20}C_{2} - {}^{20}C_{3} + \dots + {}^{20}C_{10} \end{array}$

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