JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 23)
If $$\left( {{{{3^6}} \over {{4^4}}}} \right)k$$ is the term, independent of x, in the binomial expansion of $${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$$, then k is equal to ___________.
Answer
55
Explanation
$${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$$
$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$$
$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{12 - 3r}}$$
Term independent of x $$\Rightarrow$$ 12 $$-$$ 3r = 0 $$\Rightarrow$$ r = 4
$${T_5} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^8}{\left( {12} \right)^4} = {{{3^6}} \over {{4^4}}}.\,k$$
$$\Rightarrow$$ k = 55
$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$$
$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{12 - 3r}}$$
Term independent of x $$\Rightarrow$$ 12 $$-$$ 3r = 0 $$\Rightarrow$$ r = 4
$${T_5} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^8}{\left( {12} \right)^4} = {{{3^6}} \over {{4^4}}}.\,k$$
$$\Rightarrow$$ k = 55
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