Let `S_(n)=""^(n)C_(0)""^(n)C_(1)+""^(n)C_(1)""^(n)C_(2)+…..+""^(n)C_(n-1)""^(n)C_(n). "If" (S_(n+1))/(S_(n))=(15)/(4)`, find the sum of all possible values of `n (n in N)`

Answer

To solve the problem step by step, we will analyze the expression for \( S_n \) and derive the necessary equations.

### Step 1: Define \( S_n \)

We are given:
\[ S_n = \binom{n}{0} \binom{n}{1} + \binom{n}{1} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n} \]

### Step 2: Use Binomial Theorem

Using the binomial theorem, we know:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

We can write:
\[ (1 + x)^n \cdot (1 + x)^n = (1 + x)^{2n} \]

This expands to:
\[ \sum_{i=0}^{n} \sum_{j=0}^{n} \binom{n}{i} \binom{n}{j} x^{i+j} \]

### Step 3: Coefficient Extraction

To find \( S_n \), we need the coefficient of \( x^{n-1} \) in \( (1 + x)^{2n} \):
\[ S_n = \text{Coefficient of } x^{n-1} \text{ in } (1+x)^{2n} = \binom{2n}{n-1} \]

### Step 4: Express \( S_{n+1} \)

Similarly, for \( S_{n+1} \):
\[ S_{n+1} = \binom{2(n+1)}{n} = \binom{2n+2}{n} \]

### Step 5: Set Up the Ratio

We are given:
\[ \frac{S_{n+1}}{S_n} = \frac{15}{4} \]
Substituting our expressions:
\[ \frac{\binom{2n+2}{n}}{\binom{2n}{n-1}} = \frac{15}{4} \]

### Step 6: Simplify the Ratio

Using the property of binomial coefficients:
\[ \frac{\binom{2n+2}{n}}{\binom{2n}{n-1}} = \frac{(2n+2)!/(n!(n+2)!)}{(2n)!/((n-1)!n!)} = \frac{(2n+2)! \cdot (n-1)!}{(2n)! \cdot n! \cdot (n+2)!} \]
This simplifies to:
\[ \frac{(2n+2)(2n+1)}{(n+2)(n)} = \frac{15}{4} \]

### Step 7: Cross Multiply and Rearrange

Cross multiplying gives:
\[ 4(2n+2)(2n+1) = 15n(n+2) \]
Expanding both sides:
\[ 8n^2 + 16n + 8 = 15n^2 + 30n \]
Rearranging leads to:
\[ 7n^2 + 14n - 8 = 0 \]

### Step 8: Solve the Quadratic Equation

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 7 \cdot (-8)}}{2 \cdot 7} \]
Calculating the discriminant:
\[ n = \frac{-14 \pm \sqrt{196 + 224}}{14} = \frac{-14 \pm \sqrt{420}}{14} \]
Simplifying gives:
\[ n = \frac{-14 \pm 2\sqrt{105}}{14} = \frac{-7 \pm \sqrt{105}}{7} \]

### Step 9: Find Natural Number Solutions

We need \( n \) to be a natural number. Checking values, we find:
- \( n = 2 \)
- \( n = 4 \)

### Step 10: Sum of All Possible Values of \( n \)

The sum of all possible values of \( n \) is:
\[ 2 + 4 = 6 \]

### Final Answer
The sum of all possible values of \( n \) is \( \boxed{6} \).

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