Q52 / 120 · JEE Main 2007 · Mathematics
The sum of the series
${ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+{ }^{20} \mathrm{C}_2-{ }^{20} \mathrm{C}_3+\ldots-\ldots+{ }^{20} \mathrm{C}_{10} \text { is }$
- A$-{ }^{20} \mathrm{C}_{10}$
- Correct answerB$\frac{1}{2}{ }^{20} \mathrm{C}_{10}$
- C$0$
- D${ }^{20} \mathrm{C}_{10}$
Solution
$(1+\mathrm{x})^{20}={ }^{20} \mathrm{C}_0+{ }^{20} \mathrm{C}_1 \mathrm{x}+\ldots+{ }^{20} \mathrm{C}_{10} \mathrm{x}{ }^{10}+\ldots+{ }^{20} \mathrm{C}_{20} \mathrm{x}^{20}$
put $\mathrm{x}=-1$,
$0={ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+\ldots-{ }^{20} \mathrm{C}_9+{ }^{20} \mathrm{C}_{10}-{ }^{20} \mathrm{C}_{11}+\ldots+{ }^{20} \mathrm{C}_{20}$
$0=2\left({ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+\ldots-{ }^{20} \mathrm{C}_9\right)+{ }^{20} \mathrm{C}_{10}$
$\Rightarrow{ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+\ldots+{ }^{20} \mathrm{C}_{10}=\frac{1}{2}{ }^{20} \mathrm{C}_{10}$.
Hence, the correct answer is option: 'B'.