QJEEM22Ma5759: JEE Main 2022 Mathematics - Class 11 JEEM22Ma5759 Solution: If the constant term in the expansion of is 2 k .l, where l is…

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

Given,

${\left( {3{x^2} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$

$= {{{{(3{x^8} - 2{x^7} + 5)}^{10}}} \over {{x^{50}}}}$

Now constant term in ${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}} = {x^{50}}$ term in ${(3{x^8} - 2{x^7} + 5)^{10}}$

General term in ${(3{x^8} - 2{x^7} + 5)^{10}}$ is

$= {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3{x^8})^{{n_1}}}{( - 2{x^7})^{{n_2}}}{(5)^{{n_3}}}$

$= {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n^1}}}{( - 2)^{{n_2}}}{(5)^{{n^3}}}\,.\,{x^{8{n_1} + 7{n_2}}}$

$\therefore$ Coefficient of ${x^{8{n_1} + 7{n_2}}}$ is

$= {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n_1}}}{( - 2)^{{n_2}}}{(5)^{{n_3}}}$

where ${n_1} + {n_2} + {n_3} = 0$

For coefficient of x⁵⁰ :

$8{n_1} + 7{n_2} = 50$

$\therefore$ Possible values of n₁, n₂ and n₃ are

n $_1$	n $_2$	n $_3$
1	6	3

$\therefore$ Coefficient of x⁵⁰

$= {{10!} \over {1!\,6!\,3!}}{(3)^1}{( - 2)^6}{(5)^3}$

$= {{10 \times 9 \times 8 \times 7} \over 6} \times 3 \times {5^3} \times {2^6}$

$= 5 \times 3 \times 8 \times 7 \times 3 \times {5^3} \times {2^6}$

$= 7 \times {5^4} \times {3^2} \times {2^9}$

$= {2^k}\,.\,l$

$\therefore$ $l = 7 \times {5^4} \times {3^2}$ = An odd integer

and ${2^k} = {2^9}$

$\Rightarrow k = 9$