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Equation

If 1+x n= ∑...

Question

If ${(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}$ , then the coefficient $x^{4}$ in the product of ${(1 + x)}^{n} {(1 - x)}^{n}$ is

A

C_{2}

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B

C_{4}

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C

C_{0}

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D

C_{2}^{2}

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Solution

The correct option is A $C_{2}$
$(1 + x)^{n} (1 - x)^{n}$
$= (1 - x^{2})^{n}$
Writing the general term, we get
$T_{r + 1} = (- 1)^{r}^{n} C_{r} x^{2 r}$
Hence for the coefficient of $x^{4}$
$2 r = 4$
$r = 2$
Thus the coefficient is
$^{n} C_{2}$

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Similar questions

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

, then the value of

C_{0} - C_{2} + C_{4} - C_{6} + C_{8} - C_{10} + . . .

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

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + \dots (C_{0} + C_{1} + C_{2} + \dots + C_{n - 1})

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

C_{0} + C_{2} + C_{4} + C_{6} + . . . . . . . . = ?

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

C_{0} + C_{4} + C_{8} . . . . . . . . =

{(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

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