If `t_(t)` is the rth term in the expansion of `( 1+ x)^(101)`, then what is the ratio `( t_(20))/( t_(19))` equal to

Options

A. `( 20x)/( 19)`
B. 83x
C. 19x
D. `(83x)/( 19)`

Answer

To find the ratio \( \frac{T_{20}}{T_{19}} \) in the expansion of \( (1 + x)^{101} \), we will use the formula for the \( r \)-th term in the binomial expansion, which is given by:

\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

For our case, \( n = 101 \), \( x = 1 \), and \( y = x \). Thus, the terms can be expressed as follows:

1. **Finding \( T_{20} \)**:
\[ T_{20} = \binom{101}{19} (1)^{101-19} (x)^{19} = \binom{101}{19} x^{19} \]

2. **Finding \( T_{19} \)**:
\[ T_{19} = \binom{101}{18} (1)^{101-18} (x)^{18} = \binom{101}{18} x^{18} \]

3. **Calculating the ratio \( \frac{T_{20}}{T_{19}} \)**:
\[ \frac{T_{20}}{T_{19}} = \frac{\binom{101}{19} x^{19}}{\binom{101}{18} x^{18}} \]

Simplifying this, we get:
\[ \frac{T_{20}}{T_{19}} = \frac{\binom{101}{19}}{\binom{101}{18}} \cdot x \]

4. **Using the property of binomial coefficients**:
The property states that:
\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n - r + 1}{r} \]
Applying this here:
\[ \frac{\binom{101}{19}}{\binom{101}{18}} = \frac{101 - 19 + 1}{19} = \frac{83}{19} \]

5. **Final ratio**:
Thus, substituting back into our ratio:
\[ \frac{T_{20}}{T_{19}} = \frac{83}{19} x \]

### Final Answer:
The ratio \( \frac{T_{20}}{T_{19}} \) is equal to \( \frac{83}{19} x \).