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Combination

If n>2, the...

Question

If $n > 2$ , then
$C_{1} {(a - 1)}^{2} - C_{2} {(a - 2)}^{2} + C_{3} {(a - 3)}^{2} + . . . . . . . + {(- 1)}^{n} C_{n} {(a - n)}^{2}$ is equal to

A

n a

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B

a^{2}

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C

a^{2} - 2

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D

0

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Solution

The correct option is D $a^{2}$
The given expression can be written as
$\sum_{k = 1}^{n} {(- 1)}^{k - 1} C_{k} {(a - k)}^{2}$
$= \sum_{k = 1}^{n} {(- 1)}^{k - 1} C_{k} (a^{2} - 2 a k - k^{2})$
$= a^{2} \sum_{k = 1}^{n} {(- 1)}^{k - 1} C_{k} - 2 a \sum_{k = 1}^{n} {(- 1)}^{k - 1} k C_{k} + \sum_{k = 1}^{n} {(- 1)}^{k - 1} k^{2} C_{k} . . . . . . . . (1)$
But
$\sum_{k = 1}^{n} {(- 1)}^{k - 1} C_{k} - \sum_{k = 1}^{n} {(- 1)}^{1} C_{k} = 1 - 0 = 1$
$\sum_{k = 1}^{n} {(- 1)}^{k - 1} C_{k} = - \sum_{k = 1}^{n} {(- 1)}^{k} k C_{k} = 0$
and
$\sum_{k = 1}^{n} {(- 1)}^{k - 1} k^{2} C_{k} = - \sum_{k = 1}^{n} {(- 1)}^{k} k^{2} C_{k} = 0$
Substituting these in $(1)$ , we get
$\sum_{k = 1}^{n} {(- 1)}^{k - 1} C_{k} {(a - k)}^{2} = a^{2}$

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Similar questions

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}

C_{r}

{^{n} C}_{r}

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

C_{0}, C_{1}, C_{2}, C_{3}, . . ., C_{n}

denote the binomial coefficients in the expansion of

{(1 + x)}^{n}

1^{2} . C_{1} + 2^{2} . C_{2} + 3^{2} . C_{3} + . . . + n^{2} . C_{n} =

C_{0}, C_{1}, C_{2}, C_{3}, . . ., C_{n}

denote the binomial coefficients in the expansion of

{(1 + x)}^{n}

, then prove that

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

n \geq 2

3. C_{1} - 4. C_{2} + 5. C_{3} - . . . . . . + {(- 1)}^{n - 1} (n + 2) . C_{n}

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