Let (2 x ^ 2 + 3x + 4) ^ 10 = ∑ _ r = 0 ^ 20 a _ r x ^ r . Then a _ 7 a _ 13 is equal to ______

Correct answer: 0 General term = 10 ! r _ 1 ! r _ 2 ! r _ 3 ! (2 x ^ 2 ) ^ r _ 1 (3x) ^ r _ 2 (4) ^ r _ 3 a _ 7 = 10 ! . 2 ^ 3 . 3 .4 ^ 6 3 ! 1 ! 6 ! + 10 ! 2 ^ 2 . 3 ^ 3 . 4 ^ 5 2 ! 3 ! 5 ! + 10 ! . 2 .3 ^ 5 . 4 ^ 4 1 ! . 5 ! . 4 ! + 10 ! . 3 ^ 7 . 4 ^ 3 7 ! 3 ! a _ 7 = 2 ^ 3 [ 10 ! . 2 ^ 6 . 3 .4 ^ 3 6 ! 1 ! 3 ! + 10 ! 2 ^ 5 . 3 ^ 3 . 4 ^ 2 5 ! 3 ! 2 ! + 10 ! . 2 ^ 4 . 3 ^ 5 . 4 ^ 1 4 ! . 5 ! . 1 ! + 10 ! 2 ^ 3 . 3 ^ 7 . 4 ^ 0 3 ! 7 ! ] a _ 13 = 10 ! . 2 ^ 6 . 3 .4 ^ 3 6 ! 1 ! 3 ! + 10 ! 2 ^ 5 . 3 ^ 3 . 4 ^ 2 5 ! 3 ! 2 ! + 10 ! . 2 ^ 4 . 3 ^ 5 . 4 ^ 1 4 ! . 5 ! . 1 ! + 10 ! 2 ^ 3 . 3 ^ 7 . 4 ^ 0 3 ! 7 ! a _ 7 a _ 13 = 2 ^ 3 [ 10 ! . 3 .2 ^ 12 3 ! 1 ! 6 ! + 10 ! . 3 ^ 3 . 2 ^ 9 2 ! . 3 ! . 5 ! + 10 ! . 3 ^ 5 . 2 ^ 6 1 ! . 5 ! . 4 ! + 10 ! . 3 ^ 7 . 2 ^ 3 7 ! . 3 ! 10 ! . 3 .2 ^ 12 3 ! . 1 ! . 6 ! + 10 ! . 3 ^ 3 . 2 ^ 9 2 ! . 3 ! . 5 ! + 10 ! . 3 ^ 5 . 2 ^ 6 1 ! . 5 ! . 4 ! + 10 ! . 3 ^ 7 . 2 ^ 3 7 ! . 3 ! ] = 2 ^ 3 = 8

JEE Main 20204 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 4 September · Morning Shift · Mathematics · Binomial Theorem

Let (2 x ^ 2 + 3x + 4) ^ 10 = ∑ _ r = 0 ^ 20 a _ r x ^ r . Then a _...

Let

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

Then

\dfrac{{a}_{7}}{{a}_{13}}

is equal to ______

Correct answer

Step-by-step solution

General term $= \dfrac{10 !}{{r}_{1} ! {r}_{2} ! {r}_{3} !}{\left(2{x}^{2}\right)}^{{r}_{1}}{\left(3x\right)}^{{r}_{2}}{\left(4\right)}^{{r}_{3}}$

${a}_{7} = \dfrac{10 ! . {2}^{3} . 3{.4}^{6}}{3 ! 1 ! 6 !} + \dfrac{10 ! {2}^{2} . {3}^{3} . {4}^{5}}{2 ! 3 ! 5 !} + \dfrac{10 ! . 2{.3}^{5} . {4}^{4}}{1 ! . 5 ! . 4 !} + \dfrac{10 ! . {3}^{7} . {4}^{3}}{7 ! 3 !}$

${a}_{7} = {2}^{3}\left[\dfrac{10 ! . {2}^{6} . 3{.4}^{3}}{6 ! 1 ! 3 !} + \dfrac{10 ! {2}^{5} . {3}^{3} . {4}^{2}}{5 ! 3 ! 2 !} + \dfrac{10 ! . {2}^{4} . {3}^{5} . {4}^{1}}{4 ! . 5 ! . 1 !} + \dfrac{10 ! {2}^{3} . {3}^{7} . {4}^{0}}{3 ! 7 !}\right]$

${a}_{13} = \dfrac{10 ! . {2}^{6} . 3{.4}^{3}}{6 ! 1 ! 3 !} + \dfrac{10 ! {2}^{5} . {3}^{3} . {4}^{2}}{5 ! 3 ! 2 !} + \dfrac{10 ! . {2}^{4} . {3}^{5} . {4}^{1}}{4 ! . 5 ! . 1 !} + \dfrac{10 ! {2}^{3} . {3}^{7} . {4}^{0}}{3 ! 7 !}$

$\dfrac{{a}_{7}}{{a}_{13}} = {2}^{3}\left[\dfrac{\dfrac{10 ! . 3{.2}^{12}}{3 ! 1 ! 6 !} + \dfrac{10 ! . {3}^{3} . {2}^{9}}{2 ! . 3 ! . 5 !} + \dfrac{10 ! . {3}^{5} . {2}^{6}}{1 ! . 5 ! . 4 !} + \dfrac{10 ! . {3}^{7} . {2}^{3}}{7 ! . 3 !}}{\dfrac{10 ! . 3{.2}^{12}}{3 ! . 1 ! . 6 !} + \dfrac{10 ! . {3}^{3} . {2}^{9}}{2 ! . 3 ! . 5 !} + \dfrac{10 ! . {3}^{5} . {2}^{6}}{1 ! . 5 ! . 4 !} + \dfrac{10 ! . {3}^{7} . {2}^{3}}{7 ! . 3 !}}\right] = {2}^{3} = 8$

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