If the r-th and (r+1)-th terms in the expansi | vedclass

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(B) The general term of the expansion $(p+q)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} p^{n-k} q^{k}$.Given that the $r$-th term and $(r+1)$-th term are

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(B) The general term of the expansion $(p+q)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} p^{n-k} q^{k}$.Given that the $r$-th term and $(r+1)$-th term are equal,we have $T_{r} = T_{r+1}$.$T_{r} = {}^{n}C_{r-1} p^{n-r+1} q^{r-1}$ and $T_{r+1} = {}^{n}C_{r} p^{n-r} q^{r}$.Equating them: ${}^{n}C_{r-1} p^{n-r+1} q^{r-1} = {}^{n}C_{r} p^{n-r} q^{r}$.Dividing both sides by ${}^{n}C_{r-1} p^{n-r} q^{r-1}$,we get $p = \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} q$.Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have $p = \frac{n-r+1}{r} q$.Rearranging gives $pr = (n-r+1)q = nq - rq + q = q(n+1) - rq$.Thus,$pr + rq = q(n+1)$,which implies $r(p+q) = q(n+1)$.Therefore,$\frac{q(n+1)}{r(p+q)} = 1$.

If the $r$-th and $(r+1)$-th terms in the expansion of $(p+q)^{n}$ are equal,then the value of $\frac{(n+1)q}{r(p+q)}$ is

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