If `N=1/2+1/6+1/12+1/20+1/30+………+1/156` what is the value of N?

Answer

To find the value of \( N \) given by the series \( N = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \ldots + \frac{1}{156} \), we can follow these steps:

### Step 1: Identify the Pattern
The terms in the series can be expressed in a specific form:
- \( \frac{1}{2} = \frac{1}{1 \cdot 2} \)
- \( \frac{1}{6} = \frac{1}{2 \cdot 3} \)
- \( \frac{1}{12} = \frac{1}{3 \cdot 4} \)
- \( \frac{1}{20} = \frac{1}{4 \cdot 5} \)
- \( \frac{1}{30} = \frac{1}{5 \cdot 6} \)

Continuing this pattern, we can see that the general term can be written as:
\[ \frac{1}{n(n+1)} \]
where \( n \) starts from 1 and goes up to 12 (since \( \frac{1}{156} = \frac{1}{12 \cdot 13} \)).

### Step 2: Rewrite the Terms
We can express each term \( \frac{1}{n(n+1)} \) using partial fractions:
\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]
Thus, the series can be rewritten as:
\[ N = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{12} - \frac{1}{13} \right) \]

### Step 3: Simplify the Series
When we add these terms together, we notice that most terms will cancel out:
\[ N = 1 - \frac{1}{13} \]
The remaining terms after cancellation are \( 1 \) from the first term and \( -\frac{1}{13} \) from the last term.

### Step 4: Calculate the Final Value
Now we can compute the final value of \( N \):
\[ N = 1 - \frac{1}{13} = \frac{13}{13} - \frac{1}{13} = \frac{12}{13} \]

### Conclusion
Thus, the value of \( N \) is:
\[ \boxed{\frac{12}{13}} \]