If $n > 2$ , then find the value of $C_{1} {(a - 1)}^{2} - C_{2} {(a - 2)}^{2} + C_{3} {(a - 3)}^{2} - . . . . . + {(- 1)}^{n - 1} C_{n} {(a - n)}^{2}$ where $C_{r}$ stands for ${^{n} C}_{r}$

a^{3}

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a

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\frac{a}{2}

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a^{2}

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Solution

The correct option is B $a^{2}$
Let $n = 3$
Hence the above expression reduces to
$^{3} C_{1} (a - 1)^{2} -^{3} C_{2} (a - 2)^{2} +^{3} C_{3} (a - 3)^{3}$
$= 3 (a - 1)^{2} - 3 (a - 2)^{2} + (a - 3)^{3}$
$= 3 a^{2} - 6 a + 3 - (3 a^{2} - 12 a + 12) + (a^{2} - 6 a + 9)$
$= a^{2}$
Hence answer is Option D

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Similar questions

n > 2

C_{1} {(a - 1)}^{2} - C_{2} {(a - 2)}^{2} + C_{3} {(a - 3)}^{2} + . . . . . . . + {(- 1)}^{n} C_{n} {(a - n)}^{2}

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If $^{n} C_{r} = C_{r}$ then $C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . + (C_{0} + C_{1} + . . . . . + C_{n})$ =

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