Series Sum Calculation

The problem asks for the sum of the first 20 terms of the series:

$ S = \frac{1}{5 \times 6} + \frac{1}{6 \times 7} + \frac{1}{7 \times 8} + \dots $

This is a specific type of series known as a telescoping series. We can analyze the general term of the series.

General Term Analysis

The general term of the series can be written as $\frac{1}{n(n+1)}$. Using the method of partial fractions, we can decompose this term:

$ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} $

Multiplying both sides by $n(n+1)$ gives $1 = A(n+1) + Bn$.

Setting $n=0$, we get $1 = A(1) \implies A=1$.

Setting $n=-1$, we get $1 = B(-1) \implies B=-1$.

So, the general term is:

$ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} $

Summation Steps

The series starts with the term $\frac{1}{5 \times 6}$, which corresponds to $n=5$. We need the sum of the first 20 terms. The terms correspond to $n = 5, 6, 7, \dots, (5 + 20 - 1) = 24$.

The sum $S_{20}$ can be written as:

$ S_{20} = \sum_{n=5}^{24} \frac{1}{n(n+1)} = \sum_{n=5}^{24} \left( \frac{1}{n} - \frac{1}{n+1} \right) $

Let's expand the sum to see the telescoping effect:

$ S_{20} = \left( \frac{1}{5} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{8} \right) + \dots + \left( \frac{1}{23} - \frac{1}{24} \right) + \left( \frac{1}{24} - \frac{1}{25} \right) $

Notice that the second part of each term cancels out the first part of the next term (e.g., $-\frac{1}{6}$ cancels with $+\frac{1}{6}$). This leaves only the first part of the first term and the second part of the last term:

$ S_{20} = \frac{1}{5} - \frac{1}{25} $

Final Calculation

To find the final sum, calculate the difference:

$ S_{20} = \frac{1}{5} - \frac{1}{25} = \frac{5}{25} - \frac{1}{25} = \frac{4}{25} $

Converting this fraction to a decimal:

$ S_{20} = \frac{4}{25} = 0.16 $