Question
The sum of the series $$^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - ..... + {\,^{20}}{C_{10}}$$ is
A.
$$0$$
B.
$$^{20}{C_{10}}$$
C.
$$ - ^{20}{C_{10}}$$
D.
$${\frac{1}{2}^{20}}{C_{10}}$$
Answer :
$${\frac{1}{2}^{20}}{C_{10}}$$
Solution :
We know that, $${\left( {1 + x} \right)^{20}} = {\,^{20}}{C_0} - {\,^{20}}{C_1}x + {\,^{20}}{C_2}{x^2} + ..... + {\,^{20}}{C_{10}}{x^{10}} + .....{\,^{20}}{C_{20}}{x^{20}}$$
$$\eqalign{
& {\text{Put }}x = - 1,\,\,\left( 0 \right) = {\,^{20}}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... + {\,^{20}}{C_{10}} - {\,^{20}}{C_{11}}..... + {\,^{20}}{C_{20}} \cr
& \Rightarrow \,\,0 = 2\left[ {^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - {\,^{20}}{C_9} + {\,^{20}}{C_{10}}} \right] \cr
& \Rightarrow \,{\,^{20}}{C_{10}} = 2\left[ {^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - {\,^{20}}{C_9} + {\,^{20}}{C_{10}}} \right] \cr
& \Rightarrow \,{\,^{20}}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... + {\,^{20}}{C_{10}} = \frac{1}{2}{\,^{20}}{C_{10}} \cr} $$