JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 4)
The natural number m, for which the coefficient of x in the binomial expansion of
$${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$$ is 1540, is .............
$${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$$ is 1540, is .............
Answer
13
Explanation
General term,
$${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$$
$$ \because $$ $${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$$
$$ \therefore $$ $$r = 3\,or\,19$$
$$22m - mr - 2r = 1$$
$$m = {{2r + 1} \over {22 - 5}}$$
When $$r = 3$$, $$m = {7 \over {19}} \notin N$$
When $$r = 19$$, $$m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13$$
$$ \therefore $$ $$m = 13$$
$${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$$
$$ \because $$ $${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$$
$$ \therefore $$ $$r = 3\,or\,19$$
$$22m - mr - 2r = 1$$
$$m = {{2r + 1} \over {22 - 5}}$$
When $$r = 3$$, $$m = {7 \over {19}} \notin N$$
When $$r = 19$$, $$m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13$$
$$ \therefore $$ $$m = 13$$
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