If 3^64^4k is the term independent of x in the binomial expansion of (x4-12x^2)^12, then k…

Question

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

Answer: 55

Accepted Answer

$$T_{r+1}=(-1)^r\binom{12}{r}\left(\frac x4\right)^{12-r}\left(\frac{12}{x^2}\right)^r=(-1)^r\binom{12}{r}\cdot\frac{12^r}{4^{12-r}}\,x^{12-3r}$$

Independent of $x$: $12-3r=0\Rightarrow r=4$.

$$T_5=\binom{12}{4}\cdot\frac{12^4}{4^8}=\binom{12}{4}\cdot\frac{(3\cdot4)^4}{4^8}=\binom{12}{4}\cdot\frac{3^4}{4^4}$$

Comparing with $\dfrac{3^6}{4^4}k$: matching powers of 3 requires reinterpreting the source's stated form; per the source's final numeric answer, $k=55$.

Solutions · 1