In the binomial expansion of (a - b)^n, n ge | vedclass

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(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r + 1} = {}^nC_r a^{n - r} (-b)^r$.The $5^{th}$ term is $T_5 = T_{4 + 1} = {}^nC_

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$\frac{1}{5}(n - 4)$

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(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r + 1} = {}^nC_r a^{n - r} (-b)^r$.The $5^{th}$ term is $T_5 = T_{4 + 1} = {}^nC_4 a^{n - 4} (-b)^4 = {}^nC_4 a^{n - 4} b^4$.The $6^{th}$ term is $T_6 = T_{5 + 1} = {}^nC_5 a^{n - 5} (-b)^5 = -{}^nC_5 a^{n - 5} b^5$.Given that $T_5 + T_6 = 0$,we have:${}^nC_4 a^{n - 4} b^4 - {}^nC_5 a^{n - 5} b^5 = 0$Rearranging the terms:${}^nC_4 a^{n - 4} b^4 = {}^nC_5 a^{n - 5} b^5$Dividing both sides by $a^{n - 5} b^4$:$\frac{a}{b} = \frac{{}^nC_5}{{}^nC_4}$Using the formula ${}^nC_r = \frac{n!}{r!(n - r)!}$:$\frac{a}{b} = \frac{n!}{5!(n - 5)!} 	imes \frac{4!(n - 4)!}{n!}$$\frac{a}{b} = \frac{(n - 4)!}{(n - 5)!} 	imes \frac{4!}{5!}$$\frac{a}{b} = \frac{n - 4}{5}$.

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $\frac{a}{b}$ is equal to

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