If `sum_(r=0)^(n){("^(n)C_(r-1))/('^(n)C_(r )+^(n)C_(r-1))}^(3)=(25)/(24)`, then `n` is equal to (a) 3 (b) 4 (c) 5 (d) 6

Options

A. `3`
B. `4`
C. `5`
D. `6`

Explanation

To solve the given problem, we need to evaluate the expression and find the value of \( n \) such that:

\[ \sum_{r=0}^{n} \left( \frac{nC_{r-1}}{nC_{r} + nC_{r-1}} \right)^3 = \frac{25}{24} \]

### Step 1: Rewrite the Expression
We start by rewriting the term inside the summation:

\[ \frac{nC_{r-1}}{nC_{r} + nC_{r-1}} = \frac{nC_{r-1}}{nC_{r} + nC_{r-1}} = \frac{1}{\frac{nC_{r}}{nC_{r-1}} + 1} \]

### Step 2: Simplify the Binomial Coefficients
Using the property of binomial coefficients, we know:

\[ \frac{nC_{r}}{nC_{r-1}} = \frac{n-r+1}{r} \]

So we can rewrite our expression as:

\[ \frac{nC_{r-1}}{nC_{r} + nC_{r-1}} = \frac{1}{\frac{n-r+1}{r} + 1} = \frac{r}{n+1} \]

### Step 3: Substitute Back into the Summation
Now substituting this back into our summation gives us:

\[ \sum_{r=0}^{n} \left( \frac{r}{n+1} \right)^3 = \frac{1}{(n+1)^3} \sum_{r=0}^{n} r^3 \]

### Step 4: Use the Formula for the Sum of Cubes
The formula for the sum of the cubes of the first \( n \) natural numbers is:

\[ \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

Thus, we can write:

\[ \sum_{r=0}^{n} r^3 = 0^3 + \left( \frac{n(n+1)}{2} \right)^2 = \left( \frac{n(n+1)}{2} \right)^2 \]

### Step 5: Substitute the Sum into the Equation
Substituting this back into our equation gives:

\[ \frac{1}{(n+1)^3} \cdot \left( \frac{n(n+1)}{2} \right)^2 = \frac{25}{24} \]

### Step 6: Simplify the Equation
This simplifies to:

\[ \frac{n^2(n+1)^2}{4(n+1)^3} = \frac{25}{24} \]

Cancelling \( (n+1)^2 \) from the numerator and denominator, we have:

\[ \frac{n^2}{4(n+1)} = \frac{25}{24} \]

### Step 7: Cross Multiply to Solve for \( n \)
Cross multiplying gives:

\[ 24n^2 = 100(n + 1) \]

### Step 8: Rearranging the Equation
Rearranging the equation leads to:

\[ 24n^2 - 100n - 100 = 0 \]

### Step 9: Solving the Quadratic Equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 24, b = -100, c = -100 \):

\[ n = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 24 \cdot (-100)}}{2 \cdot 24} \]

Calculating the discriminant:

\[ n = \frac{100 \pm \sqrt{10000 + 9600}}{48} \]
\[ n = \frac{100 \pm \sqrt{19600}}{48} \]
\[ n = \frac{100 \pm 140}{48} \]

Calculating the two possible values:

1. \( n = \frac{240}{48} = 5 \)
2. \( n = \frac{-40}{48} \) (not valid since \( n \) must be non-negative)

Thus, the only valid solution is:

\[ n = 5 \]

### Final Answer
The value of \( n \) is:

\[ \boxed{5} \]