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Fundamental Laws of Logarithms

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Question

Passage-2
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 $n \sum r = 1 a_{2 r - 1}$ is

A

(\frac{9^{n} - 1}{2})

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B

(\frac{3^{2 n} - 1}{4})

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C

(\frac{3^{2 n} + 1}{4})

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D

(\frac{9^{n} + 1}{2})

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Solution

The correct option is B $(\frac{3^{2 n} - 1}{4})$
$(1 + x + x^{2})^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} . . . + a_{4 n} x^{4 n}$
Substituting $x = 1$ , we get
$3^{2 n} = a_{0} + a_{1} + a_{2} . . . + a_{4 n}$ ...(i)
Substituting $x = - 1$ , we get
$1^{2 n} = a_{0} - a_{1} + a_{2} . . . + a_{4 n}$ ...(ii)
Subtracting (ii) from (i), we get
$3^{2 n} - 1 = 2 (a_{1} + a_{3} + a_{5} . . . + a_{4 n - 1})$
$\frac{3^{2 n} - 1}{2} = a_{1} + a_{3} + a_{5} . . . + a_{4 n - 1}$
Now $a_{4 n - r} = a_{r}$
$\frac{3^{2 n} - 1}{2} = a_{1} + a_{3} + a_{5} . . . + a_{2 n - 1} + (a_{2 n + 1} + a_{2 n + 3} + a_{2 n + 5} . . . + a_{4 n - 1})$
$\frac{3^{2 n} - 1}{2} = 2 (a_{1} + a_{3} + a_{5} . . . + a_{2 n - 1})$
$a_{1} + a_{3} + a_{5} . . . + a_{2 n - 1} = \frac{3^{2 n} - 1}{4}$
$\sum_{r = 1}^{n} a_{2 r - 1} = \frac{3^{2 n} - 1}{4}$

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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

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

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

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

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

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