If 36, 84, 126 are three successive binomial coefficients in the expansion of `(1+x)^(n)`, find n.

Answer

To solve the problem of finding \( n \) given that 36, 84, and 126 are three successive binomial coefficients in the expansion of \( (1+x)^n \), we can follow these steps:

### Step 1: Identify the Binomial Coefficients
Let the three successive binomial coefficients be represented as:
- \( \binom{n}{r-1} = 36 \)
- \( \binom{n}{r} = 84 \)
- \( \binom{n}{r+1} = 126 \)

### Step 2: Set Up the Ratios
Using the properties of binomial coefficients, we can set up the following ratios:
1. From \( \binom{n}{r} \) and \( \binom{n}{r-1} \):
\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{84}{36} = \frac{7}{3} \]

2. From \( \binom{n}{r+1} \) and \( \binom{n}{r} \):
\[ \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{126}{84} = \frac{3}{2} \]

### Step 3: Simplify the First Ratio
Using the formula for binomial coefficients:
\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n - r + 1}{r} \]
Setting this equal to \( \frac{7}{3} \):
\[ \frac{n - r + 1}{r} = \frac{7}{3} \]
Cross-multiplying gives:
\[ 3(n - r + 1) = 7r \]
Expanding and rearranging:
\[ 3n - 3r + 3 = 7r \implies 3n - 10r + 3 = 0 \quad \text{(Equation 1)} \]

### Step 4: Simplify the Second Ratio
Using the formula for binomial coefficients:
\[ \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n - r}{r + 1} \]
Setting this equal to \( \frac{3}{2} \):
\[ \frac{n - r}{r + 1} = \frac{3}{2} \]
Cross-multiplying gives:
\[ 2(n - r) = 3(r + 1) \]
Expanding and rearranging:
\[ 2n - 2r = 3r + 3 \implies 2n - 5r - 3 = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( 3n - 10r + 3 = 0 \)
2. \( 2n - 5r - 3 = 0 \)

We can solve these equations simultaneously. From Equation 1:
\[ 3n = 10r - 3 \implies n = \frac{10r - 3}{3} \]
Substituting \( n \) in Equation 2:
\[ 2\left(\frac{10r - 3}{3}\right) - 5r - 3 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 2(10r - 3) - 15r - 9 = 0 \]
Expanding:
\[ 20r - 6 - 15r - 9 = 0 \implies 5r - 15 = 0 \implies r = 3 \]

### Step 6: Substitute \( r \) Back to Find \( n \)
Substituting \( r = 3 \) back into the equation for \( n \):
\[ n = \frac{10(3) - 3}{3} = \frac{30 - 3}{3} = \frac{27}{3} = 9 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{9} \]