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Sum of binomial coefficients with alternate signs

If n C r = C ...

Question

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

A

n.2^n-1

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B

(n+2)2ⁿ

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C

2ⁿ

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D

(n+2)2^n-1

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Solution

The correct option is D
(n+2)2^n-1

$C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . + (C_{0} + C_{1} + . . . . . + C_{n})$ =

$n . C_{0} + (n - 1) C_{1} + (n - 2) C_{2} + . . . . . . . + C_{n} =$

$C_{0} + 2 C_{1} + 3 C_{2} + . . . . . . + n . C_{n} [∴^{n} C_{r} =^{n} C_{n - r}]$

= (n+2) $2^{n - 1} [∴ a . C_{0} + (a + d) C_{1} + (a + 2 d) C_{1} + . . . . . . + (a + (n - 1) d) C_{n} = (2 a + n d) 2^{n - 1}$ ]

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Similar questions

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

are the Binomial coefficients in the expansion

(1 + x)^{n} .

‘n’ being even, then

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . . (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1})

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

are the binomial coefficient in the expansion of

(1 + x)^{n}, n

being even, then

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . + C_{n - 1})

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

are the binomial coefficients in the expansion of

(1 + x)^{n}

. n being even, then

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . . + C_{n - 1})

{(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . . . + C_{n} x^{R}

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1}

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

denote the coefficients in the expansion of

{(1 + x)}^{n}

c_{0} c_{r} + c_{1} c_{r + 1} + c_{2} c_{r + 2} + . . . . + c_{n - r} c_{n} = \frac{| 2 n - - -}{| n - r - ---- - | n + r - ---- -}

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