In the binomial expansion of (a-b)^n, n geq 5 | vedclass

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(A) The general term in the expansion of $(a-b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} (-b)^r$. For the $5^{	ext{th}}$ term,$r=4$: $T_5 = \bi

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

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(A) The general term in the expansion of $(a-b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} (-b)^r$. For the $5^{	ext{th}}$ term,$r=4$: $T_5 = \binom{n}{4} a^{n-4} (-b)^4 = \binom{n}{4} a^{n-4} b^4$. For the $6^{	ext{th}}$ term,$r=5$: $T_6 = \binom{n}{5} a^{n-5} (-b)^5 = -\binom{n}{5} a^{n-5} b^5$. Given $T_5 + T_6 = 0$,we have $\binom{n}{4} a^{n-4} b^4 - \binom{n}{5} a^{n-5} b^5 = 0$. This implies $\binom{n}{4} a^{n-4} b^4 = \binom{n}{5} a^{n-5} b^5$. Dividing both sides by $\binom{n}{4} a^{n-5} b^4$,we get $\frac{a}{b} = \frac{\binom{n}{5}}{\binom{n}{4}}$. Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have $\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n!}{5!(n-5)!} 	imes \frac{4!(n-4)!}{n!} = \frac{n-4}{5}$. Thus,$\frac{a}{b} = \frac{n-4}{5}$.

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

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