If $n$ is an integer greater than $1$ , then $a - {^{n} C}_{1} (a - 1) + {^{n} C}_{2} (a - 2) + . . . . + {(- 1)}^{n} (a - n)$ is equal to

a

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0

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a^{2}

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2^{n}

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Solution

The correct option is A $0$
$a - {^{n} C}_{1} (a - 1) + {^{n} C}_{2} (a - 2) + . . . . + {(- 1)}^{n} (a - n)$
$= a -^{n} C_{1} a +^{n} C_{1} +^{n} C_{2} a - 2^{n} C_{2} + . . . . + (- 1)^{n} a - (- 1)^{n} n^{n} C_{n}$
$= a {C_{0} - C_{1} + C_{2} - C_{3} + . . . . . . {(- 1)}^{n} C_{n}} + {C_{1} - {2 C}_{2} + 3 C_{3} - . . . . . . {(- 1)}^{n} {n . C}_{n}}$
$= a .0 + 0 = 0$

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