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Binomial Coefficients

If 1+xn=C0+C1...

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}, t h e n C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . + C_{n}^{2}$ is equal to :

A

²ⁿC_n-1

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B

²ⁿC_n

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C

²ⁿC_n+1

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D

None

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Solution

The correct option is B
²ⁿC_n

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

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

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

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

Equating the coefficient of $x^{n}$ both sides

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . + C_{n}^{2} =^{2 n} C_{n}$

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Similar questions

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

C_{o}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . + C_{n}^{2}

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

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}$ =

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

n

' is odd,then value of

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

{(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}

, then calculated value of

C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - C_{3}^{2} + . . . (- 1)^{n} C_{n}^{2}

n = 15

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

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . + C_{n}^{2}

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