QJEEM20Maf52f: JEE Main 2020 Mathematics - Class 11 JEEM20Maf52f Solution: Let Then is equal to ______.

Note : Multinomial Theorem :

The general term of ${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$ the expansion is

${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$

where n₁ + n₂ + ..... + n_n = n

Here, in ${(2{x^2} + 3x + 4)^{10}}$ general term is

$= {{10!} \over {{n_1}!{n_2}!{n_3}!}}{(2{x^2})^{{n_1}}}{(3x)^{{n_2}}}{(4)^{{n_3}}}$

$= {{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}.{x^{2{n_1} + {n_2}}}$

$\therefore$ Coefficient of ${x^{2{n_1} + {n_2}}}$ is

${{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}$

where ${n_1} + {n_2} + {n_3} = 10$

For, Coefficient of x⁷ :
2n₁ + n₂ = 7

Possible values of n₁, n₂ and n₃ are

${n_1}$	${n_2}$	${n_3}$
3	1	6
2	3	5
1	5	4
0	7	3

$\therefore$ Coefficient of x⁷

$= {{10!} \over {3!1!6!}}{(2)^3}{(3)^1}{(4)^6} + {{10!} \over {3!1!6!}}{(2)^2}{(3)^3}{(4)^5} + {{10!} \over {1!5!4!}}{(2)^1}{(3)^5}{(4)^4} + {{10!} \over {0!7!3!}}{(2)^0}{(3)^7}{(4)^3}$

Coefficient of x¹³ = a₁₃

Here 2n₁ + n₂ = 13


possible values of n₁, n₂ and n₃ are

${n_1}$	${n_2}$	${n_3}$
6	1	3
5	3	2
4	5	1
3	7	0

$\therefore$ Coefficient of x¹³

$= {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$

$\therefore$ ${{{a_7}} \over {{a_{13}}}} = 8$