Let 2 x 2 + 3 x + 4 10 = ∑ r = 0 20 a r x r , then the value of a 12 a 8 , is

Given, 2 x 2 + 3 x + 4 10 = ∑ r = 0 20 a r x r Replacing x by 2 x , we get 8 x 2 + 6 x + 4 10 = ∑ r = 0 20 a r 2 x r ⇒ 2 10 2 x 2 + 3 x + 4 10 = ∑ r = 0 20 a r 2 r x 20 - r ⇒ 2 10 ∑ r = 0 20 a r x r = ∑ r = 0 20 a r 2 r x 20 - r Comparing ccefficient of x 8 both sides, we get 2 10 a 8 = a 12 2 12 ⇒ a 8 a 12 = 2 12 2 10 = 4

Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremLogarithmic and Exponential seriesHard2 minPYQ_2019

MathematicsHardsingle choice

Let ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} a_{r} x^{r}$ , then the value of $\frac{a _{8}}{a _{12}},$ is

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Solution:

Given, ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} a_{r} x^{r}$

Replacing $x$ by $\frac{2}{x}$ , we get

${(\frac{8}{x^{2}} + \frac{6}{x} + 4)}^{10} = \sum_{r = 0}^{20} a_{r} {(\frac{2}{x})}^{r}$

$\Rightarrow 2^{10} {(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} a_{r} 2^{r} x^{20 - r}$

$\Rightarrow 2^{10} \sum_{r = 0}^{20} a_{r} x^{r} = \sum_{r = 0}^{20} a_{r} 2^{r} x^{20 - r}$

Comparing ccefficient of $x^{8}$ both sides, we get

$2^{10} a_{8} = a_{12} 2^{12} \Rightarrow \frac{a_{8}}{a_{12}} = \frac{2^{12}}{2^{10}} = 4$

Stream:BITSATSubject:MathematicsTopic:Binomial TheoremSubtopic:Logarithmic and Exponential series

⏱ 2mℹ️ Source: PYQ_2019

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