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Proof by mathematical induction

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

A

\frac{9^{n} - 2 a_{2 n} - 1}{4}

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B

\frac{9^{n} + 2 a_{2 n} + 1}{4}

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C

\frac{9^{n} - 2 a_{2 n} + 1}{4}

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D

\frac{9^{n} + 2 a_{2 n} - 1}{4}

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Solution

The correct option is C $\frac{9^{n} - 2 a_{2 n} + 1}{4}$
$(1 + x + x^{2})^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} . . . + a_{4 n} x^{4 n}$
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)
$3^{2 n} + 1 = 2 (a_{0} + a_{2} + a_{4} . . . + a_{4 n})$
$\frac{3^{2 n} + 1}{2} = a_{0} + a_{2} + a_{4} . . . + a_{4 n}$
Now $a_{4 n - r} = a_{r}$
$\frac{3^{2 n} + 1}{2} = (a_{0} + a_{2} + a_{4} . . . + a_{2 n - 2}) + a_{2 n} + (a_{2 n + 2} + a_{2 n + 4} + a_{2 n + 6} . . . + a_{4 n})$
$\frac{3^{2 n} - 2 a_{2} + 1}{2} = 2 (a_{0} + a_{2} + a_{4} . . . + a_{4 n})$
$a_{0} + a_{2} + a_{4} . . . + a_{4 n} = \frac{3^{2 n} - 2 a_{2} + 1}{4}$
$\sum_{r = 1}^{n - 1} a_{2 r} = \frac{3^{2 n} - 2 a_{2} + 1}{4}$

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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

a_{4 n - 1}

(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}

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} = 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

n

\frac{9^{n + 1} + 4^{n + 1}}{9^{n} + 4^{n}} = 6

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