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General Term of Binomial Expansion

Consider 1+...

Question

Consider $(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}$ , where $a_{0}, a_{1}, a_{2} . . . . . . a_{4 n}$ are real numbers and n is a positive integer
The value of $a_{4 n - 1}$ is

A

2 n

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B

2 n^{2} + 4 n

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C

2 n + 3

`

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D

2 n^{2} + 3 n

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Solution

The correct option is C $2 n$
${(1 + x + x^{2})}^{2 n} = 4 n \sum r = 0 a_{r} x^{r}$
Differentiating w.r.t $x$ ,
${2 n (1 + x + x^{2})}^{2 n - 1} (1 + 2 x) = 4 n \sum r = 0 r a_{r} x^{r}$
Setting $x = 0$ , we get
$a_{1} = 2 n$

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Similar questions

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

a_{0}, a_{1}, a_{2}, . . ., a_{4 n}

are real numbers and

n

is a positive integer. The value of

a_{2}

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

a_{0}, a_{1}, a_{2} . . . . . . a_{4 n}

are real numbers and

n

is a positive integer
The correct statement is

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

a_{0}, a_{1}, a_{2} . . . . . . a_{4 n}

are real numbers and

n

is a positive integer
The value of

n - 1 \sum r = 0 a_{2 r}

Q. Passage-2
Consider

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

a_{0}, a_{1}, a_{2} . . . . . . a_{4 n}

are real numbers and

n

is a positive integer
The value of

n \sum r = 1 a_{2 r - 1}

{(1 + x + x^{2})}^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{4 n} x^{4 n}

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