If \(\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k}\) is the term, independent of x , in the binomial expansion of \(\left(\frac{\mathrm{x}}{4}-\frac{12}{\mathrm{x}^{2}}\right)^{12}\), then k is equal to ____.

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Binomial Theorem

(\frac{3 ^{6}}{4 ^{4}}) k

is the term, independent of x , in the binomial expansion of

(\frac{x}{4} - \frac{12}{x ^{2}})^{12}

, then k is equal to ____.

Answer: 55

The general term of

(\frac{x}{4} - \frac{12}{x ^{2}})^{12}

T_{r + 1} = (r 12) (\frac{x}{4})^{12 - r} (- \frac{12}{x ^{2}})^{r} = (r 12) \frac{( - 12 ) ^{r}}{4 ^{12 - r}} x^{12 - 3 r}

. The term independent of

x

needs

12 - 3 r = 0

, so

r = 4

, giving

T_{5} = (4 12) \frac{1 2 ^{4}}{4 ^{8}}

. Since

1 2^{4} = 3^{4} \cdot 4^{4}

, this is

(4 12) \frac{3 ^{4}}{4 ^{4}} = 495 \cdot \frac{3 ^{4}}{4 ^{4}}

. Written as

\frac{3 ^{6}}{4 ^{4}} k

, we get

k = \frac{495}{3 ^{2}} = 55

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer