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Question: If \[p+q=1\] then \[\sum{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}},r\in \left[ r=0,n \right]}\] is equ...

If p+q=1p+q=1 then ∑r.nCrprqn−r,r∈[r=0,n]\sum{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}},r\in \left[ r=0,n \right]} is equal to
1. 11
2. np
3. npq
4. 00

Explanation

Solution

Hint : We are given a summation formula in terms of r. Here, we will expand the given summation equation using the rules of permutations and combinations and binomial theorem. On further simplifying the expansion, we can obtain the required answer.

Complete step-by-step answer :
We start with the expression given to us which is
∑r.nCrprqn−r,r∈[r=0,n]\sum{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}},r\in \left[ r=0,n \right]}
Now this can be written as
∑r=0nr.nCrprqn−r\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}
Now opening the summation we get
∑r=0nr.nCrprqn−r=0.nC0p0qn−0+1.nC1p1qn−1+2.nC2p2qn−2+3.nC3p3qn−3+........(n−1)nCn−1pn−1q1+n.nCnpnq0\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}={{0.}^{n}}{{C}_{0}}{{p}^{0}}{{q}^{n-0}}+{{1.}^{n}}{{C}_{1}}{{p}^{1}}{{q}^{n-1}}+{{2.}^{n}}{{C}_{2}}{{p}^{2}}{{q}^{n-2}}+{{3.}^{n}}{{C}_{3}}{{p}^{3}}{{q}^{n-3}}+........{{(n-1)}^{n}}{{C}_{n-1}}{{p}^{n-1}}{{q}^{1}}+n{{.}^{n}}{{C}_{n}}{{p}^{n}}{{q}^{0}}
Now simplifying this we get
∑r=0nr.nCrprqn−r=n!1!(n−1)!p1qn−1+2n!2!(n−2)!p2qn−2+3n!3!(n−3)!p3qn−3+........(n−1)n!n−1!(n−n+1)!pn−1q1+nn!n!(n−n)!pnq0\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}=\dfrac{n!}{1!\left( n-1 \right)!}{{p}^{1}}{{q}^{n-1}}+2\dfrac{n!}{2!\left( n-2 \right)!}{{p}^{2}}{{q}^{n-2}}+3\dfrac{n!}{3!\left( n-3 \right)!}{{p}^{3}}{{q}^{n-3}}+........(n-1)\dfrac{n!}{n-1!\left( n-n+1 \right)!}{{p}^{n-1}}{{q}^{1}}+n\dfrac{n!}{n!\left( n-n \right)!}{{p}^{n}}{{q}^{0}}
Solving this further
∑r=0nr.nCrprqn−r=npqn−1+n(n−1)p2qn−2+n(n−1)(n−2)p3qn−3+........\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}=np{{q}^{n-1}}+n(n-1){{p}^{2}}{{q}^{n-2}}+n(n-1)(n-2){{p}^{3}}{{q}^{n-3}}+........
Now we can also write this as after taking out np out as common from all terms to be
∑r=0nr.nCrprqn−r=np(qn−1+(n−1)pqn−2+(n−1)(n−2)p2qn−3+........)\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}=np({{q}^{n-1}}+(n-1)p{{q}^{n-2}}+(n-1)(n-2){{p}^{2}}{{q}^{n-3}}+........)
We know that the expression inside the bracket now can be written as
∑r=0nr.nCrprqn−r=np(q+p)n−1\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}=np{{(q+p)}^{n-1}}
Now in the question it is given to us that the sum of p and q will always no matter what will be equal to one which also gives us that
∑r=0nr.nCrprqn−r=np(1)n−1\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}=np{{(1)}^{n-1}}
Now everything raised to one always gives us one no matter what therefore we now get that
∑r=0nr.nCrprqn−r=np\sum\limits_{r=0}^{n}{r{{.}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}=np
Now this is the answer which was asked to us in the above question therefore we can say that the answer of this question is option 2 that is np.
So, the correct answer is “Option 2”.

Note : A common mistake here is that some might forget to put the r in the substitution of the terms in the opening of expression. So one must try to avoid making that mistake also after opening the terms instead of solving the expression step by step since it won’t be possible because we have three variables we must try to deduce the expression in such a way that we can use the logic of the sum of p and q being one to simplify it and get the answer of this question since you can’t further solve it without that.