If (3^6/4^4)k is the term, independent of x, in the binomial expansion of
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08/02/2022 3:02 pm
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If \(\Big(\frac{3^6}{4^4}\Big)\)k is the term, independent of x, in the binomial expansion of \(\Big(\frac{x}{4}  \frac{12}{x^2}\Big)^{12}\), then k is equal to .............
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08/02/2022 3:13 pm
\(\Big(\frac{x}{4}  \frac{12}{x^2}\Big)^{12}\)
T_{r+1} = (1)^{r}. \(^{12}C_r(\frac{x}{4})^{12r}(\frac{12}{x^2})^r\)
T_{r+1} = (1)^{r}. \(^{12}C_r(\frac{1}{4})^{12r}(12)^r.(x)^{123r}\)
Term independent of x ⇒ 12 – 3r = 0 ⇒ r = 4
T_{5} = (1)^{4}.\(^{12}C_4(\frac{1}{4})^8(12)^4\)
= \(\frac{3^6}{4^4}\).k
⇒ k = 55
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