If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ and ' $n$ ' is odd,then value of $C_{0}^{2} - C_{1}^{2} + C_{2}^{2} + . . . + {(- 1)}^{^{n}} C_{n}^{2}$ is

^{2 n} C_{n + 1}

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0

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{(- 1)}^{\frac{n}{2}} \frac{n!}{(\frac{n}{2})! (\frac{n}{2}!)}

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^{2 n} C_{n - 2}

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Solution

The correct option is A $0$
$(^{n} C_{0}^{2} -^{n} C_{1}^{2} +^{n} C_{2}^{2} . .)$
$= (^{n} C_{0} -^{n} C_{1} +^{n} C_{2} . .) \times (^{n} C_{0} -^{n} C_{1} +^{n} C_{2} . .)$
$= [(1 + x)^{n} (1 - x)^{n}]_{x = 1}$
$= (1 - x^{2})_{x = 1}^{n}$
$= 0$
Hence the above expression yields 0.

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Similar questions

Q. If

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

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

n = 15

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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 :

Q. Prove that according as n is odd or even the value of

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

(- 1)^{n / 2} \frac{n!}{(n / 2)! (n / 2)!}

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If (1+x)n=C0+C1x+C2x2+...+Cnxn and 'n' is odd,then value of C20−C21+C22+...+(−1)nC2n is

The correct option is A 0(nC20−nC21+nC22..)=(nC0−nC1+nC2..)×(nC0−nC1+nC2..)=[(1+x)n(1−x)n]x=1=(1−x2)nx=1=0Hence the above expression yields 0.

If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ and ' $n$ ' is odd,then value of $C_{0}^{2} - C_{1}^{2} + C_{2}^{2} + . . . + {(- 1)}^{^{n}} C_{n}^{2}$ is

The correct option is A $0$
$(^{n} C_{0}^{2} -^{n} C_{1}^{2} +^{n} C_{2}^{2} . .)$
$= (^{n} C_{0} -^{n} C_{1} +^{n} C_{2} . .) \times (^{n} C_{0} -^{n} C_{1} +^{n} C_{2} . .)$
$= [(1 + x)^{n} (1 - x)^{n}]_{x = 1}$
$= (1 - x^{2})_{x = 1}^{n}$
$= 0$
Hence the above expression yields 0.