A variant of the newton's binomial theorem sum with extra conditions on p and q

Question

I am trying to compute this sum which at first glance seems to be easy but always either fail or get a result that is self-contradictory.

the sum : $$\mathscr{S} = \sum_{k=0}^nk{n \choose k}p^kq^{n-k}$$ where $p$ , $q$ ∈ $\Bbb{R^{*}+}$ and $p + q = 1$

My last try : ok so first of all because the first term is equal to zero I'm gonna change the start point from 0 to 1

$$\mathscr{S} = \sum_{k=1}^nk{n \choose k}p^kq^{n-k}$$

and then simplifiy $k{n \choose k}$

${n \choose k} = \frac{n!}{k!(n-k)!}$

$k{n \choose k} = \frac{n!}{k!(n-k)!} = \frac{n!}{(k-1)!(n-k)!}$

plug the result above in $\mathscr{S}$

$$\mathscr{S} = \sum_{k=1}^n\frac{n!}{(k-1)!(n-k)!}p^kq^{n-k}$$

and since $n$ is a free variable I can pull it out of the sum

$\mathscr{S} = n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}p^kq^{n-k} =n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}p^kq^{n-k} = n\sum_{k=1}^nk{n-1 \choose k-1}p^kq^{n-k}$

and since $p$ and $q$ are just like $n$ two free variables I'll pull one $p$ and one $q$ too from the sum and I'll get this

$\mathscr{S} = npq\sum_{k=1}^n{n-1 \choose k-1}p^{k-1}q^{n-(k-1)}$

now I'll change the start point to 0 and I get this :

$\mathscr{S} = npq\sum_{k=0}^n{n-1 \choose k}p^{k}q^{n-k} = pq\sum_{k=0}^n\frac{n!}{k!(n-k-1)!}p^{k}q^{n-k}$

now I divide and multiply the inside of the sum by $(n-k)$

$\mathscr{S} = pq\sum_{k=0}^n(n-k)\frac{n!}{k!(n-k-1)!(n-k)}p^{k}q^{n-k} =pq\sum_{k=0}^n(n-k)\frac{n!}{k!(n-k)!}p^{k}q^{n-k} = pq\sum_{k=0}^n$$(n-k)$${n \choose k}p^kq^{n-k}$

by rearranging I get this : $\mathscr{S} = pq\sum_{k=0}^nn{n \choose k}p^kq^{n-k}-pq\sum_{k=0}^nk{n \choose k}p^kq^{n-k} = npq\sum_{k=0}^n{n \choose k}p^kq^{n-k} -pq\mathscr{S}$

and according to the newton's binomial theorem we have that $$\sum_{k=0}^n{n \choose k}p^kq^{n-k} = (p+q)^n=1^n=1$$

so I concluded that $\mathscr{S}$ is basically equal to $\frac{npq}{1+pq}$

But for $n = 2$ and $q=p=1/2$

$$\sum_{k=0}^2k{2 \choose k}(0.5)^k(0.5)^{2-k} = 1 \neq \frac{2*0.5*0.5}{1+0.5*0.5} = 0.4$$

so I've made a mistake somewhere but can't find where exactly

please help me find my mistake and solve this problem.

thank you

Answer 1

When p and q are pulled out, it should be:

$$ \mathscr{S} = n\frac p q\sum_{k=1}^n{n-1 \choose k-1}p^{k-1}q^{n-(k-1)} $$

Answer 2

Hint:

Try operating on both sides of $$(p+q)^n=\sum_{k=0}^{n}\binom{n}{k}p^kq^{n-k}$$ with $p\frac{\partial}{\partial p}$ to obtain a formula.

Answer 3

Another way is to recal the binomial identy $$ k\left( \begin{gathered} n \\ k \\ \end{gathered} \right) = n\left( \begin{gathered} n - 1 \\ k - 1 \\ \end{gathered} \right) $$ so that $$ \begin{gathered} S = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {k\left( \begin{gathered} n \\ k \\ \end{gathered} \right)p^{\,k} q^{\,n - k} } = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {n\left( \begin{gathered} n - 1 \\ k - 1 \\ \end{gathered} \right)p^{\,k} q^{\,n - k} } = \hfill \\ = n\,p\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered} n - 1 \\ k - 1 \\ \end{gathered} \right)p^{\,k - 1} q^{\,\left( {n - 1} \right) - \left( {k - 1} \right)} } = n\,p\sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered} n - 1 \\ j \\ \end{gathered} \right)p^{\,j} q^{\,\left( {n - 1} \right) - j} } = \hfill \\ = n\,p\left( {p + q} \right)^{\,n - 1} \hfill \\ \end{gathered} $$