TS EAMCET · Maths · Binomial Theorem
Numerically greatest term in the expansion of \((3 x-4 y)^{23}\) when \(x=\frac{1}{6}\) and \(y=\frac{1}{8}\) is
- A \(\frac{{ }^{23} \mathrm{C}_{11}}{6^{23}}\)
- B \({ }^{23} \mathrm{C}_{11}\left(\frac{8}{6}\right)^{23}\)
- C \({ }^{23} \mathrm{C}_{11}\left(\frac{6}{8}\right)^{23}\)
- D \({ }^{23} \mathrm{C}_{11}\left(\frac{1}{2}\right)^{23}\)
Answer & Solution
Correct Answer
(D) \({ }^{23} \mathrm{C}_{11}\left(\frac{1}{2}\right)^{23}\)
Step-by-step Solution
Detailed explanation
\( \text{Let } A=3x, B=-4y, n=23. \) \( \left| \frac{B}{A} \right| = \left| \frac{-4y}{3x} \right| = \frac{4y}{3x} = \frac{4(\frac{1}{8})}{3(\frac{1}{6})} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \)…
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