If the first three terms in the binomial expansion of `(1+bx)^n` in ascending powers of x are 1,6x and `16x^2` respectively then b + n =

Options

A. A) `28/3`
B. B) `15/2`
C. C) `29/3`
D. D) `17/3`

Answer

To solve the problem, we need to analyze the binomial expansion of \((1 + bx)^n\) and find the values of \(b\) and \(n\) based on the given terms.

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The binomial expansion of \((1 + bx)^n\) is given by:
\[ (1 + bx)^n = \binom{n}{0} (1)^{n} (bx)^{0} + \binom{n}{1} (1)^{n-1} (bx)^{1} + \binom{n}{2} (1)^{n-2} (bx)^{2} + \ldots \]
This simplifies to:
\[ = 1 + n(bx) + \frac{n(n-1)}{2}(bx)^2 + \ldots \]

2. **Identifying the Terms**:
From the problem, we know that the first three terms are:
- First term: \(1\)
- Second term: \(6x\)
- Third term: \(16x^2\)

Therefore, we can equate:
- \(n \cdot b = 6\) (from the coefficient of \(x\))
- \(\frac{n(n-1)}{2} b^2 = 16\) (from the coefficient of \(x^2\))

3. **Setting Up Equations**:
From the first equation:
\[ n \cdot b = 6 \quad \text{(1)} \]
From the second equation:
\[ \frac{n(n-1)}{2} b^2 = 16 \quad \text{(2)} \]

4. **Express \(b\) in terms of \(n\)**:
From equation (1), we can express \(b\) as:
\[ b = \frac{6}{n} \]

5. **Substituting \(b\) into Equation (2)**:
Substitute \(b\) into equation (2):
\[ \frac{n(n-1)}{2} \left(\frac{6}{n}\right)^2 = 16 \]
Simplifying this gives:
\[ \frac{n(n-1)}{2} \cdot \frac{36}{n^2} = 16 \]
\[ \frac{36(n-1)}{2n} = 16 \]
\[ 18(n-1) = 16n \]
\[ 18n - 18 = 16n \]
\[ 2n = 18 \]
\[ n = 9 \]

6. **Finding \(b\)**:
Now substitute \(n = 9\) back into equation (1) to find \(b\):
\[ 9b = 6 \implies b = \frac{6}{9} = \frac{2}{3} \]

7. **Calculating \(b + n\)**:
Now we can find \(b + n\):
\[ b + n = \frac{2}{3} + 9 = \frac{2}{3} + \frac{27}{3} = \frac{29}{3} \]

### Final Answer:
Thus, the value of \(b + n\) is \(\frac{29}{3}\).