Et 2 x2-3 x-410-r-020 ar xr - Then the value of a g-a11 is equal toA- 1B- 2 C- 1-2

The correct option is B 2(2x^2+3x+4)^10= ∑_r=0^20a_rx^r Replace x by 2x, we get nbsp; ( 8x^2+6x+4 )^10=∑_r=0^20a_r·2^rx^r ⇒2^10(4+3x+2x^2)^10x^20=∑_r=0^20a_ ...

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Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

et 2 x2+3 x+4...

Question

Let $(2 x^{2} + 3 x + 4)^{10} = 20 \sum r = 0 a_{r} x^{r} .$ Then the value of $\frac{a_{9}}{a_{11}}$ is equal to

1

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\frac{1}{2}

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Solution

The correct option is B $2$
$(2 x^{2} + 3 x + 4)^{10} = 20 \sum r = 0 a_{r} x^{r}$
Replace $x$ by $\frac{2}{x},$ we get
${(\frac{8}{x^{2}} + \frac{6}{x} + 4)}^{10} = 20 \sum r = 0 \frac{a_{r} \cdot 2^{r}}{x^{r}}$
$\Rightarrow \frac{2^{10} (4 + 3 x + 2 x^{2})^{10}}{x^{20}} = 20 \sum r = 0 a_{r} 2^{r} x^{- r}$
$\Rightarrow 2^{10} 20 \sum r = 0 a_{r} x^{r} = 20 \sum r = 0 a_{r} 2^{r} x^{20 - r}$
Comparing coefficient of $x^{9}$ both sides, we get
$2^{10} \times a_{9} = a_{11} \cdot 2^{11} \Rightarrow \frac{a_{9}}{a_{11}} = 2$

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Similar questions

Q. Let

(2 x^{2} + 3 x + 4)^{10} = 20 \sum r = 0 a_{r} x^{r} .

Then the value of

\frac{a_{9}}{a_{11}}

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Let (2x2+3x+4)10=20∑r=0arxr. Then the value of a9a11 is equal to

The correct option is B 2(2x2+3x+4)10=20∑r=0arxr Replace x by 2x, we get (8x2+6x+4)10=20∑r=0ar⋅2rxr ⇒210(4+3x+2x2)10x20=20∑r=0ar2rx−r ⇒21020∑r=0arxr=20∑r=0ar2rx20−r Comparing coefficient of x9 both sides, we get 210×a9=a11⋅211⇒a9a11=2

Let $(2 x^{2} + 3 x + 4)^{10} = 20 \sum r = 0 a_{r} x^{r} .$ Then the value of $\frac{a_{9}}{a_{11}}$ is equal to