For `beta ne 0`, if the coefficient of `x^(3)` in the binomial expansion of `(1 + betax)^(6)` and the coefficient of `x^(4)` in the binomial expansion of `(1 - betax)^(8)` are equal, then the value of `beta` is

Options

A. `2//7`
B. `-2//7`
C. `-1//7`
D. `1//7`

Answer

To find the value of \(\beta\) such that the coefficient of \(x^3\) in the binomial expansion of \((1 + \beta x)^6\) is equal to the coefficient of \(x^4\) in the binomial expansion of \((1 - \beta x)^8\), we can follow these steps:

### Step 1: Find the coefficient of \(x^3\) in \((1 + \beta x)^6\)

The general term in the binomial expansion of \((1 + \beta x)^n\) is given by:
\[ T_r = \binom{n}{r} (1)^{n-r} (\beta x)^r = \binom{n}{r} \beta^r x^r \]

For \(n = 6\) and \(r = 3\):
\[ T_3 = \binom{6}{3} \beta^3 x^3 \]

Calculating \(\binom{6}{3}\):
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

Thus, the coefficient of \(x^3\) is:
\[ 20 \beta^3 \]

### Step 2: Find the coefficient of \(x^4\) in \((1 - \beta x)^8\)

Using the same formula for the binomial expansion, for \(n = 8\) and \(r = 4\):
\[ T_4 = \binom{8}{4} (-\beta x)^4 = \binom{8}{4} (-\beta)^4 x^4 \]

Calculating \(\binom{8}{4}\):
\[ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

Thus, the coefficient of \(x^4\) is:
\[ 70 \beta^4 \]

### Step 3: Set the coefficients equal

We set the coefficients from the two expansions equal to each other:
\[ 20 \beta^3 = 70 \beta^4 \]

### Step 4: Simplify the equation

Dividing both sides by \(\beta^3\) (since \(\beta \neq 0\)):
\[ 20 = 70 \beta \]

### Step 5: Solve for \(\beta\)

Now, we can solve for \(\beta\):
\[ \beta = \frac{20}{70} = \frac{2}{7} \]

### Conclusion

Thus, the value of \(\beta\) is:
\[ \boxed{\frac{2}{7}} \]