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

Options

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

Explanation

To find the ratio of the 20th term \( T_{20} \) to the 19th term \( T_{19} \) 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:

\[ T_r = \binom{n}{r-1} x^{n-(r-1)} \]

where \( n \) is the exponent of the binomial, and \( r \) is the term number.

### Step 1: Write the general term for the expansion
For the expansion of \( (1+x)^{101} \), the \( r \)-th term is:

\[ T_r = \binom{101}{r-1} x^{101-(r-1)} = \binom{101}{r-1} x^{102-r} \]

### Step 2: Find the 20th term \( T_{20} \)
Substituting \( r = 20 \):

\[ T_{20} = \binom{101}{20-1} x^{102-20} = \binom{101}{19} x^{82} \]

### Step 3: Find the 19th term \( T_{19} \)
Substituting \( r = 19 \):

\[ T_{19} = \binom{101}{19-1} x^{102-19} = \binom{101}{18} x^{83} \]

### Step 4: Calculate the ratio \( \frac{T_{20}}{T_{19}} \)
Now, we can find the ratio of the 20th term to the 19th term:

\[ \frac{T_{20}}{T_{19}} = \frac{\binom{101}{19} x^{82}}{\binom{101}{18} x^{83}} \]

### Step 5: Simplify the ratio
This can be simplified as follows:

\[ \frac{T_{20}}{T_{19}} = \frac{\binom{101}{19}}{\binom{101}{18}} \cdot \frac{x^{82}}{x^{83}} = \frac{\binom{101}{19}}{\binom{101}{18}} \cdot \frac{1}{x} \]

Using the property of binomial coefficients:

\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} \]

we have:

\[ \frac{\binom{101}{19}}{\binom{101}{18}} = \frac{101 - 19 + 1}{19} = \frac{83}{19} \]

Thus, we can write:

\[ \frac{T_{20}}{T_{19}} = \frac{83}{19} \cdot \frac{1}{x} = \frac{83}{19x} \]

### Final Answer
The ratio \( \frac{T_{20}}{T_{19}} \) is:

\[ \frac{T_{20}}{T_{19}} = \frac{83}{19x} \]