If n - N and if 1 - 4x - 4x2n - -r - 02narxr - where ao-a1-a2-a2n are real numbers- The value of 2-r - 1n a2r - 1 - is-

The correct option is A 9^n 1 Solution : #160; (1+4x+4x^2)^n=(1+2x)^2n (1+2x)^2n=^2nC_0+^2nC_1(2x)+...........^2nC_2n(2x)^2n put x = 1 ...

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Factorisation Using Algebraic Identities

If n ∈ N an...

Question

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

9^{n} - 1

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9^{n} + 1

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9^{n} - 2

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9^{n} + 2

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Solution

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Similar questions

n \in N

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

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

2 \sum_{r = 0}^{n} a_{2 r}

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}

n \sum r = 0 a_{2 r}

n

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

a_{r}

a_{2 n - 1}

n

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

a_{r}

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

n

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

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