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Derivative from First Principle

If n ∈ N an...

Question

If $n \in N$ and if $(1 + 4 x + 4 x^{2})^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r}$ , where $a_{0}, a_{1}, a_{1}, a_{2}, . . . . . . . . ., a_{2 n}$ are real numbers.
The value of $2 \sum_{r = 0}^{n} a_{2 r}$ , is

A

9^{n} - 1

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B

9^{n} + 1

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C

9^{n} - 2

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D

9^{n} + 2

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Solution

The correct option is B $9^{n} + 1$
Let $f (x) = (1 + 4 x x + 4 x^{2})^{n}$ $= \sum_{r = 0}^{2 n} a_{r} x_{r}$
Now, $2 \sum_{r = 0}^{n} a_{2 r} = f (1) + f (- 1$
So, $2 \sum_{r = 0}^{n} a_{2 r} = 9^{n} + 1$

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Similar questions

n \in N

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

a_{o}, a_{1}, a_{2}, . . . . . . ., a_{2 n}

are real numbers. The value of

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

n \in N

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

a_{o}, a_{1}, a_{2}, . . . . . . ., a_{2 n}

are real numbers. The value of

n \sum r = 0 a_{2 r}

n

is the positive integer and if

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

a_{r}

are real numbers. The value of

a_{2 n - 1}

a_{r} > 0; \forall r, n \in N

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

are in A.P, then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

n

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

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