JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 3)
If the coefficients of x−2 and x−4 in the expansion of $${\left( {{x^{{1 \over 3}}} + {1 \over {2{x^{{1 \over 3}}}}}} \right)^{18}},\left( {x > 0} \right),$$ are m and n respectively, then $${m \over n}$$ is equal to :
182
$${4 \over 5}$$
$${5 \over 4}$$
27
Explanation
Tr+1 = 18Cr $${\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}$$ . $${\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}$$
= 18Cr $${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}$$
For coefficient of x$$-$$2,
$${{18 - 2r} \over 3}$$ = $$-$$2
$$ \Rightarrow $$ r = 12
$$ \therefore $$ Coefficient of x$$-$$2 is (m) = 18C12 $${\left( {{1 \over 2}} \right)^{12}}$$
For coefficient of x$$-$$4,
$${{18 - 2r} \over 3}$$ = $$-$$ 4
$$ \Rightarrow $$ r = 15
$$ \therefore $$ Coefficient of x$$-$$4 is (n) = 18C15 $$\left( {{1 \over {2}}} \right)$$15
$$ \therefore $$ $${m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}$$
= $${{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}$$
= 182
= 18Cr $${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}$$
For coefficient of x$$-$$2,
$${{18 - 2r} \over 3}$$ = $$-$$2
$$ \Rightarrow $$ r = 12
$$ \therefore $$ Coefficient of x$$-$$2 is (m) = 18C12 $${\left( {{1 \over 2}} \right)^{12}}$$
For coefficient of x$$-$$4,
$${{18 - 2r} \over 3}$$ = $$-$$ 4
$$ \Rightarrow $$ r = 15
$$ \therefore $$ Coefficient of x$$-$$4 is (n) = 18C15 $$\left( {{1 \over {2}}} \right)$$15
$$ \therefore $$ $${m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}$$
= $${{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}$$
= 182
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