JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 15)
Let the coefficients of third, fourth and fifth terms in the expansion of $${\left( {x + {a \over {{x^2}}}} \right)^n},x \ne 0$$, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________.
Answer
4
Explanation
$${T_{r + 1}} = {n_{C_r}}{x^{n - r}}.{\left( {{a \over {{x^2}}}} \right)^r}$$
$$ = {}^n{C_r}{a^r}{x^{n - 3r}}$$
$${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$$, $${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$$, $${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$$
Now, $${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}$$
$$ \Rightarrow a(n - 2) = 2$$ .......... (i)
and $${{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}$$
$$ \Rightarrow a(n - 3) = {3 \over 2}$$ ........ (ii)
by (i) and (ii) $$n = 6,\,a = {1 \over 2}$$
for term independent of 'x'
$$n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2$$
$${T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4$$
$$ = {}^n{C_r}{a^r}{x^{n - 3r}}$$
$${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$$, $${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$$, $${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$$
Now, $${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}$$
$$ \Rightarrow a(n - 2) = 2$$ .......... (i)
and $${{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}$$
$$ \Rightarrow a(n - 3) = {3 \over 2}$$ ........ (ii)
by (i) and (ii) $$n = 6,\,a = {1 \over 2}$$
for term independent of 'x'
$$n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2$$
$${T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4$$
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