If `T_(2)//T_(3)` in the expansion of `(a+b)^(n) and T_(3)//T_(4)` in the expansion of `(a+b)^(n+3)` are equal, then n =

Options

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

Answer

To solve the problem step by step, we need to find the value of \( n \) such that \( \frac{T_2}{T_3} \) in the expansion of \( (a+b)^n \) is equal to \( \frac{T_3}{T_4} \) in the expansion of \( (a+b)^{n+3} \).

### Step 1: Write the general term for the binomial expansion
The general term \( T_{r+1} \) in the expansion of \( (a+b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

### Step 2: Find \( T_2 \) and \( T_3 \) in the expansion of \( (a+b)^n \)
For \( T_2 \) (where \( r = 1 \)):
\[ T_2 = \binom{n}{1} a^{n-1} b^1 = n a^{n-1} b \]

For \( T_3 \) (where \( r = 2 \)):
\[ T_3 = \binom{n}{2} a^{n-2} b^2 = \frac{n(n-1)}{2} a^{n-2} b^2 \]

### Step 3: Write the ratio \( \frac{T_2}{T_3} \)
Now, we can write the ratio:
\[ \frac{T_2}{T_3} = \frac{n a^{n-1} b}{\frac{n(n-1)}{2} a^{n-2} b^2} = \frac{2n a^{n-1} b}{n(n-1) a^{n-2} b^2} \]
This simplifies to:
\[ \frac{T_2}{T_3} = \frac{2a}{(n-1)b} \]

### Step 4: Find \( T_3 \) and \( T_4 \) in the expansion of \( (a+b)^{n+3} \)
For the expansion \( (a+b)^{n+3} \):

For \( T_3 \) (where \( r = 2 \)):
\[ T_3 = \binom{n+3}{2} a^{(n+3)-2} b^2 = \frac{(n+3)(n+2)}{2} a^{n+1} b^2 \]

For \( T_4 \) (where \( r = 3 \)):
\[ T_4 = \binom{n+3}{3} a^{(n+3)-3} b^3 = \frac{(n+3)(n+2)(n+1)}{6} a^{n} b^3 \]

### Step 5: Write the ratio \( \frac{T_3}{T_4} \)
Now, we can write the ratio:
\[ \frac{T_3}{T_4} = \frac{\frac{(n+3)(n+2)}{2} a^{n+1} b^2}{\frac{(n+3)(n+2)(n+1)}{6} a^{n} b^3} \]
This simplifies to:
\[ \frac{T_3}{T_4} = \frac{3(n+3)(n+2)a^{n+1} b^2}{(n+3)(n+2)(n+1)a^{n} b^3} = \frac{3a}{(n+1)b} \]

### Step 6: Set the two ratios equal
Now we set the two ratios equal to each other:
\[ \frac{2a}{(n-1)b} = \frac{3a}{(n+1)b} \]

### Step 7: Simplify and solve for \( n \)
Cancelling \( a \) and \( b \) (assuming \( a \neq 0 \) and \( b \neq 0 \)):
\[ \frac{2}{n-1} = \frac{3}{n+1} \]
Cross-multiplying gives:
\[ 2(n+1) = 3(n-1) \]
Expanding both sides:
\[ 2n + 2 = 3n - 3 \]
Rearranging gives:
\[ 2 + 3 = 3n - 2n \implies n = 5 \]

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{5} \]