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 $T_{r+1} = ^nC_r a^{n-r} (-b)^r$.For the $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4}

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$\frac{n - 4}{5}$

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(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} (-b)^r$.For the $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4} (-b)^4 = ^nC_4 a^{n-4} b^4$.For the $6^{th}$ term $(r=5)$: $T_6 = ^nC_5 a^{n-5} (-b)^5 = -^nC_5 a^{n-5} b^5$.Given $T_5 + T_6 = 0$,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 $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:$\frac{a}{b} = \frac{n-5+1}{5} = \frac{n-4}{5}$.

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

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