The natural number $m$, for which the coefficient of $x$ in the binomial expansion of $\left(x^{m}+\frac{1}{x^{2}}\right)^{22}$ is 1540, is ______

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Binomial Theorem

The natural number

m

, for which the coefficient of

x

in the binomial expansion of

(x^{m} + \frac{1}{x ^{2}})^{22}

is 1540, is ______

Answer: 13

The general term of

(x^{m} + \frac{1}{x ^{2}})^{22}

(r 22) x^{m (22 - r) - 2 r}

, with coefficient

(r 22)

. The required coefficient is

1540

, and

(r 22) = 1540

only for

r = 3

r = 19

, since

(3 22) = (19 22) = 1540

. The power of

x

must equal

1

: taking

r = 19

gives

m (22 - 19) - 2 (19) = 3 m - 38 = 1

, so

m = 13

, whereas

r = 3

would need

19 m = 7

, which has no natural-number solution. Hence

m = 13

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Integer answer