JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 3)
Let $${\left( {2{x^2} + 3x + 4} \right)^{10}} = \sum\limits_{r = 0}^{20} {{a_r}{x^r}} $$
Then $${{{a_7}} \over {{a_{13}}}}$$ is equal to ______.
Then $${{{a_7}} \over {{a_{13}}}}$$ is equal to ______.
Answer
8
Explanation
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 + ..... + nn = 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
$$ \therefore $$ 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
$$ \therefore $$ Coefficient of x13
$$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$$
$$ \therefore $$ $${{{a_7}} \over {{a_{13}}}} = 8$$
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 + ..... + nn = 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
$$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$$
$$ \therefore $$ $${{{a_7}} \over {{a_{13}}}} = 8$$
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