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

Correct answer: B. 4 Given,& (2x^ 2 + 3x + 4)10 = ∑ _ r = 0 ^ 20 a _ r x^ r Replacing x by (2)/(x) , we get ((8)/(x2) + (6)/(x) + 4)10 = ∑ _ r = 0 ^ 20 a _ r ((2)/(x))r ⇒ 2 ^ 10 (2x^ 2 + 3x + 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 ⇒ (a8)/(a12) = (212)/(210) = 4

BITSAT 2019

Exam: BITSAT
Year: 2019
Subject: Mathematics
Chapter: Binomial Theorem

BITSAT 2019 · Mathematics · Binomial Theorem

Let (2x^ 2 + 3x + 4)10 = ∑ _ r = 0 ^ 20 a _ r x^ r , then the value...

Let

\left(2x^{2} + 3x + 4\right)10 = \mathop{∑}_{r = 0}^{20}{a}_{r}x^{r}

, then the value of

\frac{a_{8}}{a_{12}},

Options

A $2$
B $4$
C $8$
D $16$

Correct answer

B. $4$

Step-by-step solution

Given,& $\left(2x^{2} + 3x + 4\right)10 = \mathop{∑}_{r = 0}^{20}{a}_{r}x^{r}$

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

\left(\dfrac{8}{x2} + \dfrac{6}{x} + 4\right)10 = \mathop{∑}_{r = 0}^{20}{a}_{r}\left(\dfrac{2}{x}\right)r $$ ⇒ {2}^{10}\left(2x^{2} + 3x + 4\right)10 = \mathop{∑}_{r = 0}^{20}{a}_{r}{2}^{r}x^{20 - r} $$ ⇒ {2}^{10}\mathop{∑}_{r = 0}^{20}{a}_{r}x^{r} = \mathop{∑}_{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} ⇒ \dfrac{a8}{a12} = \dfrac{212}{210} = 4$

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