Solveeit Logo
Login

Question

Question: If \({{P}_{n}}\) denotes the product of the binomial coefficients in the expansion of \({{\left( 1+x...

If Pn{{P}_{n}} denotes the product of the binomial coefficients in the expansion of (1+x)n{{\left( 1+x \right)}^{n}} ,then Pn+1Pn\dfrac{{{P}_{n+1}}}{{{P}_{n}}} equals
(a) (n+1)nn!\dfrac{{{\left( n+1 \right)}^{n}}}{n!}
(b) nnn!\dfrac{{{n}^{n}}}{n!}
(c) (n+1)n(n+1)!\dfrac{{{\left( n+1 \right)}^{n}}}{\left( n+1 \right)!}
(d) (n+1)n+1(n+1)!\dfrac{{{\left( n+1 \right)}^{n+1}}}{\left( n+1 \right)!}

Explanation

Solution

In order to solve this problem, we need to find the coefficients of the binomial coefficients of the expansion.
The binomial theorem states that,
(a+b)n=(nC0)an+(nC1)an−1b+(nC2)an−2b2+......+(nCn−1)abn−1+(nCn)bn{{\left( a+b \right)}^{n}}=\left( {}^{n}{{C}_{0}} \right){{a}^{n}}+\left( {}^{n}{{C}_{1}} \right){{a}^{n-1}}b+\left( {}^{n}{{C}_{2}} \right){{a}^{n-2}}{{b}^{2}}+......+\left( {}^{n}{{C}_{n-1}} \right)a{{b}^{n-1}}+\left( {}^{n}{{C}_{n}} \right){{b}^{n}}, where nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . We can see that the irrespective of the expression the coefficients remain the same.

Complete step-by-step answer :
We are given the expression and we need to find the binomial coefficients of the expansion.
The binomial theorem states that,
(a+b)n=(nC0)an+(nC1)an−1b+(nC2)an−2b2+......+(nCn−1)abn−1+(nCn)bn{{\left( a+b \right)}^{n}}=\left( {}^{n}{{C}_{0}} \right){{a}^{n}}+\left( {}^{n}{{C}_{1}} \right){{a}^{n-1}}b+\left( {}^{n}{{C}_{2}} \right){{a}^{n-2}}{{b}^{2}}+......+\left( {}^{n}{{C}_{n-1}} \right)a{{b}^{n-1}}+\left( {}^{n}{{C}_{n}} \right){{b}^{n}}
Where nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} .
We are given that the product of the coefficients of the binomial of (1+x)n{{\left( 1+x \right)}^{n}} be Pn{{P}_{n}} .
We can see that irrespective of the expression the coefficients remain the same.
Therefore, Pn=(nC0)(nC1)(nC2)......(nCn).................(i){{P}_{n}}=\left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)......\left( {}^{n}{{C}_{n}} \right).................(i)
Therefore, the term Pn+1=(n+1C0)(n+1C1)(n+2C2)......(n+1Cn+1).............................(ii){{P}_{n+1}}=\left( {}^{n+1}{{C}_{0}} \right)\left( {}^{n+1}{{C}_{1}} \right)\left( {}^{n+2}{{C}_{2}} \right)......\left( {}^{n+1}{{C}_{n+1}} \right).............................(ii) .
Solving equation (i) we get,
Pn=(nC0)(nC1)(nC2)......(nCn) =(n!n!)(n!(n−1)!)(n!2(n−2)!)......(n!n!) =1×(n1!)×(n(n−1)2!)×(n(n−1)(n−2)3!).........(n!n!)......................(iii)\begin{aligned} & {{P}_{n}}=\left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)......\left( {}^{n}{{C}_{n}} \right) \\\ & =\left( \dfrac{n!}{n!} \right)\left( \dfrac{n!}{\left( n-1 \right)!} \right)\left( \dfrac{n!}{2\left( n-2 \right)!} \right)......\left( \dfrac{n!}{n!} \right) \\\ & =1\times \left( \dfrac{n}{1!} \right)\times \left( \dfrac{n\left( n-1 \right)}{2!} \right)\times \left( \dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!} \right).........\left( \dfrac{n!}{n!} \right)......................(iii) \end{aligned}
We can write the value Pn+1{{P}_{n+1}} just by replacing (n) by (n+1).
Therefore,
Pn+1=(n+11!)×((n+1)(n)2!)×((n+1)(n)(n−1)3!).........((n+1)!(n+1)!)......................(iv){{P}_{n+1}}=\left( \dfrac{n+1}{1!} \right)\times \left( \dfrac{\left( n+1 \right)\left( n \right)}{2!} \right)\times \left( \dfrac{\left( n+1 \right)\left( n \right)\left( n-1 \right)}{3!} \right).........\left( \dfrac{\left( n+1 \right)!}{\left( n+1 \right)!} \right)......................(iv)
Dividing the equation (iv) by (iii), we get,
Pn+1Pn=(n+11!)×((n+1)(n)2!)×((n+1)(n)(n−1)3!).........((n+1)!(n+1)!)1×(n1!)×(n(n−1)2!)×(n(n−1)(n−2)3!).........(n!n!)\dfrac{{{P}_{n+1}}}{{{P}_{n}}}=\dfrac{\left( \dfrac{n+1}{1!} \right)\times \left( \dfrac{\left( n+1 \right)\left( n \right)}{2!} \right)\times \left( \dfrac{\left( n+1 \right)\left( n \right)\left( n-1 \right)}{3!} \right).........\left( \dfrac{\left( n+1 \right)!}{\left( n+1 \right)!} \right)}{1\times \left( \dfrac{n}{1!} \right)\times \left( \dfrac{n\left( n-1 \right)}{2!} \right)\times \left( \dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!} \right).........\left( \dfrac{n!}{n!} \right)}
As we can see that the factorial in the denominators gets cancelled except the last terms
Therefore, by solving we get,
Pn+1Pn=(n+1)(n+1)(n)(n+1)(n)(n−1).....(n+1)!(n+1)!(n)(n)(n−1)(n)(n−1)(n−2).....(n!)\dfrac{{{P}_{n+1}}}{{{P}_{n}}}=\dfrac{\dfrac{\left( n+1 \right)\left( n+1 \right)\left( n \right)\left( n+1 \right)\left( n \right)\left( n-1 \right).....\left( n+1 \right)!}{\left( n+1 \right)!}}{\left( n \right)\left( n \right)\left( n-1 \right)\left( n \right)\left( n-1 \right)\left( n-2 \right).....\left( n! \right)}
There is (n+1) terms (n+1) times till the end in the numerator, whereas there are n terms in the denominator n times.
Similarly, there is one n less than the number of terms that is (n+1) times. Therefore, there are n terms n times in the numerator.
So we can show that with the same analogy that (n-1) terms are repeated (n-1) terms.
In the same way in the denominator, there are (n-1) terms repeated (n-1) times and so on.
Hence, the expression becomes,
Pn+1Pn=(n+1)n+1(n)n(n−1)n−1......1(n)n(n−1)n−1(n−2)n−2......1(n+1)!\dfrac{{{P}_{n+1}}}{{{P}_{n}}}=\dfrac{{{\left( n+1 \right)}^{n+1}}{{\left( n \right)}^{n}}{{\left( n-1 \right)}^{n-1}}......1}{{{\left( n \right)}^{n}}{{\left( n-1 \right)}^{n-1}}{{\left( n-2 \right)}^{n-2}}......1\left( n+1 \right)!}
Solving this further we get,
Pn+1Pn=(n+1)n+1(n)n(n−1)n−1......1(n)n(n−1)n−1(n−2)n−2......1(n+1)! =(n+1)n+1(n+1)! =(n+1)nn!\begin{aligned} & \dfrac{{{P}_{n+1}}}{{{P}_{n}}}=\dfrac{{{\left( n+1 \right)}^{n+1}}{{\left( n \right)}^{n}}{{\left( n-1 \right)}^{n-1}}......1}{{{\left( n \right)}^{n}}{{\left( n-1 \right)}^{n-1}}{{\left( n-2 \right)}^{n-2}}......1\left( n+1 \right)!} \\\ & =\dfrac{{{\left( n+1 \right)}^{n+1}}}{\left( n+1 \right)!} \\\ & =\dfrac{{{\left( n+1 \right)}^{n}}}{n!} \end{aligned}
Hence, the correct option is (a).

Note : We need to understand the analogy of (n+1) appearing (n+1) times and (n) appearing (n) times and so on. There is a similar case in the denominator starting from (n) repeating (n) times, followed by (n-1) appearing (n-1) times and so on. This makes the expression simplified. Also initially while cancelling the factorials in the denominator there is one term in the upper half which does not get cancelled.