If the coefficient of `rth` term and `(r+1)^(th)` term in the expansion of `(1+x)^(20)` are in ratio `1:2`, then `r` is equal to

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( r^{th} \) term and the \( (r+1)^{th} \) term in the expansion of \( (1+x)^{20} \) are in the ratio \( 1:2 \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (1+x)^{20} \) is given by:
\[ T_{r+1} = \binom{20}{r} x^r \]
and the \( (r+1)^{th} \) term is:
\[ T_{r+2} = \binom{20}{r+1} x^{r+1} \]

2. **Set Up the Ratio**:
We know that the ratio of the coefficients of the \( r^{th} \) term to the \( (r+1)^{th} \) term is given by:
\[ \frac{\binom{20}{r}}{\binom{20}{r+1}} = \frac{1}{2} \]

3. **Use the Property of Binomial Coefficients**:
The relationship between consecutive binomial coefficients is:
\[ \binom{n}{r} = \frac{n-r}{r+1} \binom{n}{r+1} \]
Therefore, we can write:
\[ \frac{\binom{20}{r}}{\binom{20}{r+1}} = \frac{20 - r}{r + 1} \]
Setting this equal to \( \frac{1}{2} \):
\[ \frac{20 - r}{r + 1} = \frac{1}{2} \]

4. **Cross Multiply**:
Cross multiplying gives us:
\[ 2(20 - r) = 1(r + 1) \]
Simplifying this:
\[ 40 - 2r = r + 1 \]

5. **Rearrange the Equation**:
Rearranging the equation to isolate \( r \):
\[ 40 - 1 = r + 2r \]
\[ 39 = 3r \]

6. **Solve for \( r \)**:
Dividing both sides by 3:
\[ r = \frac{39}{3} = 13 \]

Thus, the value of \( r \) is \( 13 \).

### Final Answer:
\[ r = 13 \]