If `(1+ax)^(n)=1+10x+40x^(2)+….` then value of `(a+n)/(a-n)` is

Options

A. `5//7`
B. `7//3`
C. `-5//7`
D. `-7//3`

Answer

To solve the problem, we start with the equation given:

\[ (1 + ax)^n = 1 + 10x + 40x^2 + \ldots \]

### Step 1: Identify the coefficients
From the expansion of \((1 + ax)^n\), we can express it using the binomial theorem:

\[ (1 + ax)^n = \sum_{r=0}^{n} \binom{n}{r} (ax)^r = \sum_{r=0}^{n} \binom{n}{r} a^r x^r \]

### Step 2: Compare coefficients
We need to compare the coefficients of \(x\) and \(x^2\) from both sides of the equation.

- The coefficient of \(x\) is given by \(\binom{n}{1} a = 10\).
- The coefficient of \(x^2\) is given by \(\binom{n}{2} a^2 = 40\).

### Step 3: Set up equations
From the coefficients, we can set up the following equations:

1. \(\binom{n}{1} a = 10 \implies n a = 10\) (Equation 1)
2. \(\binom{n}{2} a^2 = 40 \implies \frac{n(n-1)}{2} a^2 = 40\) (Equation 2)

### Step 4: Solve Equation 1 for \(a\)
From Equation 1, we can express \(a\) in terms of \(n\):

\[ a = \frac{10}{n} \]

### Step 5: Substitute \(a\) into Equation 2
Substituting \(a\) into Equation 2 gives:

\[ \frac{n(n-1)}{2} \left(\frac{10}{n}\right)^2 = 40 \]

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

\[ \frac{n(n-1)}{2} \cdot \frac{100}{n^2} = 40 \]

Multiplying both sides by \(2n^2\):

\[ n(n-1) \cdot 100 = 80n^2 \]

### Step 7: Rearranging the equation
Rearranging gives:

\[ 100n^2 - 100n = 80n^2 \]

\[ 20n^2 - 100n = 0 \]

### Step 8: Factor the equation
Factoring out \(20n\):

\[ 20n(n - 5) = 0 \]

### Step 9: Solve for \(n\)
This gives us two solutions:

1. \(n = 0\) (not valid in this context)
2. \(n = 5\)

### Step 10: Find \(a\)
Substituting \(n = 5\) back into Equation 1:

\[ a = \frac{10}{5} = 2 \]

### Step 11: Calculate \(\frac{a+n}{a-n}\)
Now we can calculate:

\[ \frac{a+n}{a-n} = \frac{2 + 5}{2 - 5} = \frac{7}{-3} = -\frac{7}{3} \]

### Final Answer
Thus, the value of \(\frac{a+n}{a-n}\) is:

\[ \boxed{-\frac{7}{3}} \]