In the expansion of (1+x)^n,the coefficients | vedclass

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(D) In the expansion of $(1+x)^n$,the general term,i.e.,$(r+1)$-th term is given by $T_{r+1} = { }^n C_r x^r$.The coefficient of the $p$-th term is ${

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$n+1$

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(D) In the expansion of $(1+x)^n$,the general term,i.e.,$(r+1)$-th term is given by $T_{r+1} = { }^n C_r x^r$.The coefficient of the $p$-th term is ${ }^n C_{p-1}$. Given that this coefficient is $p$,we have $p = { }^n C_{p-1}$.The coefficient of the $(p+1)$-th term is ${ }^n C_p$. Given that this coefficient is $q$,we have $q = { }^n C_p$.Now,consider the ratio $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}}$.Using the formula $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$,we get:$\frac{q}{p} = \frac{n-p+1}{p}$.Cross-multiplying,we get $qp = p(n-p+1)$.Wait,let us re-evaluate the given condition: coefficients are $p$ and $q$.Actually,the problem states the coefficients are $p$ and $q$. Let's use the property $\frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.Since $q = { }^n C_p$ and $p = { }^n C_{p-1}$,we have $\frac{q}{p} = \frac{n-p+1}{p}$.This implies $q = n-p+1$,which simplifies to $p+q = n+1$.

In the expansion of $(1+x)^n$,the coefficients of $p$-th and $(p+1)$-th terms are respectively $p$ and $q$. Then $p+q=$

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