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Question: If two consecutive terms in the expansion of \({{\left( p+q \right)}^{n}}\) are equal, where n is a ...

If two consecutive terms in the expansion of (p+q)n{{\left( p+q \right)}^{n}} are equal, where n is a positive integer, then (n+1)qp+q\dfrac{\left( n+1 \right)q}{p+q} is:
(a)A rational number
(b)A positive integer
(c)A negative integer
(d)An integer

Explanation

Solution

Hint: First apply binomial theorem to find any two consecutive coefficients. By that condition, find the relation between q, p, r, n. Now substitute this relation into the given number. Check the type of number. This type is the required result in the question.

Complete step-by-step answer:
Given condition in the question is written inform of:
Consecutive terms of expansion (p+q)n{{\left( p+q \right)}^{n}} are equals. Let us assume that rth,(r+1)th{{r}^{th}},{{\left( r+1 \right)}^{th}} terms are those consecutive terms which are given to be as equal in value.
By binomial theorem we can say the expansion to be:
(p+q)n=∑k=0nnCkpkqn−k{{\left( p+q \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}}{{p}^{k}}{{q}^{n-k}}
From above expansion we can say the rth{{r}^{th}} term to be as:
Let us assume the rth{{r}^{th}}term to be denoted by ‘a’.
By substituting k=rk=r , we get value of variable ‘a’ as:
a=nCrprqn−ra={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}
From the binomial theorem we can say the (r+1)th{{\left( r+1 \right)}^{th}} term to be as:
Let us assume the (r+1)th{{\left( r+1 \right)}^{th}}term to be denoted by b.
By substituting k=r+1k=r+1 , we get the value of variable ‘b’ as:
b=nCr+1pr+1qn−(r+1)b={}^{n}{{C}_{r+1}}{{p}^{r+1}}{{q}^{n-\left( r+1 \right)}}
By simplifying above equation, we get the value of variable ‘b’ as:
b=nCr+1pr+1qn−r−1b={}^{n}{{C}_{r+1}}{{p}^{r+1}}{{q}^{n-r-1}}
By condition given between a, b we can say the equation as:
a=ba=b
By substituting their values, we get equation to be as:
nCrprqn−r=nCr+1pr+1qn−r−1{}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}={}^{n}{{C}_{r+1}}{{p}^{r+1}}{{q}^{n-r-1}}
By simplifying, combining terms , we get the equation in form:
qn−rqn−r−1=nCr+1nCr.pr+1pr\dfrac{{{q}^{n-r}}}{{{q}^{n-r-1}}}=\dfrac{{}^{n}{{C}_{r+1}}}{{}^{n}{C}_{r}}.\dfrac{{{p}^{r+1}}}{{{p}^{r}}}
Combinations:- it is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter in combination you can select items in any order. Formula is given by nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
By expanding and simplifying, we get the equation as:
qp=n!(n−r)!r!n!(r+1)!(n−r−1)!=n−rr+1\dfrac{q}{p}=\dfrac{n!\left( n-r \right)!r!}{n!\left( r+1 \right)!\left( n-r-1 \right)!}=\dfrac{n-r}{r+1} ……………………. (1)
Given number for which we need the type is give:
(n+1)qp+q\dfrac{\left( n+1 \right)q}{p+q}
By taking common from denominator and cancel it, we get :
n+1pq+1\dfrac{n+1}{\dfrac{p}{q}+1}
By substituting equation (1) , we get it as follows below:
n+1(r+1n−r)+1=n+1r+1+n−rn−r=(n+1)(n−r)n+1\dfrac{n+1}{\left( \dfrac{r+1}{n-r} \right)+1}=\dfrac{n+1}{\dfrac{r+1+n-r}{n-r}}=\dfrac{\left( n+1 \right)\left( n-r \right)}{n+1}
By cancelling the common terms, we get the equation
(n+1)qp+q=n−r\dfrac{\left( n+1 \right)q}{p+q}=n-r
As n is always greater than r, we say n−rn-r is a positive integer.
Therefore option (b) is the correct answer for this question.

Note: Be careful with the formula of binomial theorem, combination because the whole answer depends on it. Students forget to write r! terms in the denominator. Do not make that mistake as it might make the whole result go wrong, while converting numbers we took a common just to generate pq\dfrac{p}{q} term. It is an important idea.