The sum of the series : `1 + (1)/(1 + 2) + (1)/(1 + 2 + 3)+`.... up to 10 terms, is

Options

A. `(22)/(13)`
B. `(18)/(11)`
C. `(20)/(11)`
D. `(16)/(9)`

Answer

To solve the problem of finding the sum of the series \( S = 1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \ldots \) up to 10 terms, we can follow these steps:

### Step 1: Identify the nth term of the series
The nth term of the series can be expressed as:
\[ T_n = \frac{1}{1 + 2 + 3 + \ldots + n} \]
The sum of the first n natural numbers is given by the formula:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus, we can rewrite the nth term as:
\[ T_n = \frac{1}{\frac{n(n + 1)}{2}} = \frac{2}{n(n + 1)} \]

### Step 2: Write the sum of the first 10 terms
Now, we need to find the sum of the first 10 terms:
\[ S_{10} = T_1 + T_2 + T_3 + \ldots + T_{10} \]
Substituting for \( T_n \):
\[ S_{10} = \sum_{n=1}^{10} \frac{2}{n(n + 1)} \]

### Step 3: Simplify the summation
We can simplify \( \frac{2}{n(n + 1)} \) using partial fractions:
\[ \frac{2}{n(n + 1)} = \frac{2}{n} - \frac{2}{n + 1} \]
Thus, we can rewrite the sum:
\[ S_{10} = \sum_{n=1}^{10} \left( \frac{2}{n} - \frac{2}{n + 1} \right) \]

### Step 4: Evaluate the summation
This series is telescoping, which means that many terms will cancel out:
\[ S_{10} = \left( \frac{2}{1} - \frac{2}{2} \right) + \left( \frac{2}{2} - \frac{2}{3} \right) + \left( \frac{2}{3} - \frac{2}{4} \right) + \ldots + \left( \frac{2}{10} - \frac{2}{11} \right) \]
After cancellation, we are left with:
\[ S_{10} = 2 - \frac{2}{11} = 2 \cdot \frac{11}{11} - \frac{2}{11} = \frac{22}{11} - \frac{2}{11} = \frac{20}{11} \]

### Final Result
Thus, the sum of the series up to 10 terms is:
\[ \boxed{\frac{20}{11}} \]