The sum of the series
`(.^(101)C_(1))/(.^(101)C_(0)) + (2..^(101)C_(2))/(.^(101)C_(1)) + (3..^(101)C_(3))/(.^(101)C_(2)) + "….." + (101..^(101)C_(101))/(.^(101)C_(100))` is `"____"`.

Explanation

To solve the given series, we will break it down step by step.

### Given Series:
\[ S = \frac{{^{101}C_1}}{{^{101}C_0}} + \frac{{2 \cdot ^{101}C_2}}{{^{101}C_1}} + \frac{{3 \cdot ^{101}C_3}}{{^{101}C_2}} + \ldots + \frac{{101 \cdot ^{101}C_{101}}}{{^{101}C_{100}}} \]

### Step 1: Simplifying Each Term
Each term of the series can be rewritten as:
\[ \frac{{r \cdot ^{101}C_r}}{{^{101}C_{r-1}}} \]
Using the property of binomial coefficients, we know:
\[ ^{n}C_{r} = \frac{n!}{r!(n-r)!} \]
Thus, we can express:
\[ \frac{{^{101}C_r}}{{^{101}C_{r-1}}} = \frac{^{101}C_r}{\frac{101!}{(r-1)!(101-r+1)!}} = \frac{101! \cdot (r-1)! \cdot (101-r)!}{r! \cdot (101-r)!} = \frac{101}{r} \]
So, we can rewrite the term as:
\[ \frac{r \cdot ^{101}C_r}{^{101}C_{r-1}} = r \cdot \frac{^{101}C_r}{^{101}C_{r-1}} = r \cdot \frac{101}{r} = 101 \]

### Step 2: Summing the Series
Now, we can sum the series:
\[ S = 101 + 101 + 101 + \ldots + 101 \quad \text{(for r = 1 to 101)} \]
There are 101 terms in this series.

### Step 3: Final Calculation
Thus, the sum \( S \) becomes:
\[ S = 101 \times 101 = 10201 \]

### Final Answer:
The sum of the series is:
\[ \boxed{10201} \]