Let `t_(n) ` denote the `n^(th)` term in a binomial expansion. If `(t_(6))/(t_(5))` in the expansion of `(a+b)^(n+4)` and `(t_(5))/(t_(4))` in the expansion of `(a+b)^(n)` are equal, then n is

Options

A. 9
B. 11
C. 13
D. 15

Answer

To solve the problem, we need to find the value of \( n \) given the ratios of terms in two binomial expansions. Let’s break it down step by step.

### Step 1: Understand the General Term of Binomial Expansion
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^r b^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

### Step 2: Write the Terms for the First Expansion
For the expansion of \( (a + b)^{n+4} \):
- The 6th term \( T_6 \) corresponds to \( r = 5 \):
\[ T_6 = \binom{n+4}{5} a^5 b^{(n+4)-5} = \binom{n+4}{5} a^5 b^{n-1} \]
- The 5th term \( T_5 \) corresponds to \( r = 4 \):
\[ T_5 = \binom{n+4}{4} a^4 b^{(n+4)-4} = \binom{n+4}{4} a^4 b^{n} \]

### Step 3: Write the Terms for the Second Expansion
For the expansion of \( (a + b)^{n} \):
- The 5th term \( T_5 \) corresponds to \( r = 4 \):
\[ T_5 = \binom{n}{4} a^4 b^{(n)-4} = \binom{n}{4} a^4 b^{n-4} \]
- The 4th term \( T_4 \) corresponds to \( r = 3 \):
\[ T_4 = \binom{n}{3} a^3 b^{(n)-3} = \binom{n}{3} a^3 b^{n-3} \]

### Step 4: Set Up the Ratios
We know that:
\[ \frac{T_6}{T_5} = \frac{T_5}{T_4} \]
Substituting the expressions we found:
\[ \frac{\binom{n+4}{5} a^5 b^{n-1}}{\binom{n+4}{4} a^4 b^{n}} = \frac{\binom{n}{4} a^4 b^{n-4}}{\binom{n}{3} a^3 b^{n-3}} \]

### Step 5: Simplify the Ratios
The left-hand side simplifies to:
\[ \frac{\binom{n+4}{5}}{\binom{n+4}{4}} \cdot \frac{a^5}{a^4} \cdot \frac{b^{n-1}}{b^n} = \frac{\binom{n+4}{5}}{\binom{n+4}{4}} \cdot a \cdot \frac{1}{b} \]
The right-hand side simplifies to:
\[ \frac{\binom{n}{4}}{\binom{n}{3}} \cdot \frac{a^4}{a^3} \cdot \frac{b^{n-4}}{b^{n-3}} = \frac{\binom{n}{4}}{\binom{n}{3}} \cdot a \cdot \frac{1}{b} \]

### Step 6: Set the Simplified Ratios Equal
Now we equate the simplified ratios:
\[ \frac{\binom{n+4}{5}}{\binom{n+4}{4}} = \frac{\binom{n}{4}}{\binom{n}{3}} \]

### Step 7: Use the Binomial Coefficient Formula
Using the property of binomial coefficients:
\[ \frac{\binom{n+4}{5}}{\binom{n+4}{4}} = \frac{n+4-5+1}{5} = \frac{n}{5} \]
\[ \frac{\binom{n}{4}}{\binom{n}{3}} = \frac{n-4+1}{4} = \frac{n-3}{4} \]

### Step 8: Set Up the Equation
Now we have:
\[ \frac{n}{5} = \frac{n-3}{4} \]

### Step 9: Cross-Multiply and Solve for \( n \)
Cross-multiplying gives:
\[ 4n = 5(n - 3) \]
Expanding this:
\[ 4n = 5n - 15 \]
Rearranging gives:
\[ 15 = 5n - 4n \implies n = 15 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{15} \]