MCQ
The sum $^\text{n}\text{C}_0+^\text{n}\text{C}_1+^\text{n}\text{C}_2+.....+^\text{n}\text{C}_\text{n}$ is equal to
- ✓$\frac{2.4.6......2\text{n}}{\text{n!}}$
- B$\text{n}^\text{n}$
- C$\text{n!}$
- D$3^\text{n}$
We know, $\text{(x+a)}^\text{n}=\text{nc}_0\text{x}^\text{n}+\text{nc}_1\text{x}^\text{n-1}\text{a}+\text{nc}_2\text{x}^\text{n-2}\text{a}^2+.....+\text{nc}_\text{n}\text{a}^\text{n}$
Let $x = 1, a = 1$, then,
$2^\text{n}=\text{nc}_0+\text{nc}_2+.....+\text{nc}_\text{n}$
$\text{nc}_0+\text{nc}_1+\text{nc}_2+....+\text{nc}_\text{n}=2^\text{n}$
$=\frac{2.1\times2.2\times2.3....\times2.\text{n}}{1\times2\times3...\times\text{n}}=\frac{2.4.6....2\text{n}}{\text{n!}}$
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