If n - N and if 1 - 4x - 4x2n - -r - 02narXr - - where ao-a1-a2-a2n are real numbers- The value of -r - 0n a2r- is

The correct option is E None of the aboveSolution (1+4x+4x^2)^n = ∑_r = 0^2na_rx^r (1+2x)^2n = ^2nC_0 +^2nC_1(2x)+—–^2nC_2n(2x)^2n Put x = 1 ...

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Division of a Polynomial by a Monomial

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 $n \sum r = 0 a_{2 r}$ , 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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None of the above

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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}

a_{2 n - 1}

n

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

a_{r}

a_{2 n - 1}

n

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

X = {4^{n} - 3 n - 1; n \in R}

Y = {9 (n - 1); n \in N}

X \cap Y =

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