Question

What is the value of \[\sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} \]?
A. \[n + 1\]
B. \[\dfrac{n}{2}\]
C. \[n + 2\]
D. None of these

Answer 1

Hint: First, simplify the given function by rearranging the terms. Then, simplify the function by applying the combination formula in the terms of \[n\], and \[r\] . Then, solve the summation function. In the end, simplify the sum by using the formula of the sum of first \[n\] natural numbers and get the required answer.



Formula Used:The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
The sum of first \[n\] natural numbers: \[1 + 2 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}\]



Complete step by step solution:The given expression is \[\sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} \].
Let’s solve the above expression.
Rearrange the terms.
\[\sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{1 + \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}}}}} \]
Solve the right-hand side by applying the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
\[\sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{1 + \dfrac{{\left( {\dfrac{{n!}}{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}} \right)}}{{\left( {\dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right)}}}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{1 + \dfrac{{\left( {\dfrac{{n!}}{{\left( {r + 1} \right)r!\left( {n - r - 1} \right)!}}} \right)}}{{\left( {\dfrac{{n!}}{{r!\left( {n - r} \right)\left( {n - r - 1} \right)!}}} \right)}}}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{1 + \dfrac{{\left( {\dfrac{1}{{\left( {r + 1} \right)}}} \right)}}{{\left( {\dfrac{1}{{\left( {n - r} \right)}}} \right)}}}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{1 + \dfrac{{n - r}}{{r + 1}}}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{\dfrac{{r + 1 + n - r}}{{r + 1}}}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{1}{{\left( {\dfrac{{n + 1}}{{r + 1}}} \right)}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \sum\limits_{r = 0}^{n - 1} {\dfrac{{r + 1}}{{n + 1}}} \]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \dfrac{1}{{n + 1}}\sum\limits_{r = 0}^{n - 1} {\left( {r + 1} \right)} \]
Now solve the summation function.
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \dfrac{1}{{n + 1}}\left[ {1 + 2 + 3 + ... + n} \right]\]
Use the formula of the sum of first \[n\] natural numbers: \[1 + 2 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}\].
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \dfrac{1}{{n + 1}}\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]\]
\[ \Rightarrow \sum\limits_{r = 0}^{n - 1} {\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}}} = \dfrac{n}{2}\]



Option ‘B’ is correct



Note: Students often get confused between the combination and permutation formulas.
Remember the following formulas:
The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
The permutation formula: \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Factorial: \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]