To find the sums and values requested, let's break down each part of the problem step-by-step.
### Sum of the Series 1 + 1/3 + 1/6 + 1/10 + ... up to 20 Terms
We need to find the sum of the first 20 terms of the series:
[tex]\[ S = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \ldots \][/tex]
Let's analyze the pattern. Each term in the series can be written in the form:
[tex]\[ \frac{1}{\frac{n(n+1)}{2}} = \frac{2}{n(n+1)} \][/tex]
where [tex]\( n \)[/tex] is the term's index.
The sum of the first 20 terms is obtained by adding each term up:
[tex]\[ S = \sum_{n=1}^{20} \frac{2}{n(n+1)} \][/tex]
### Value of [tex]\( 1 - \left[ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{2009 \times 2010} \right] \)[/tex]
Here, we need to find the series:
[tex]\[ \sum_{n=1}^{2009} \frac{1}{n(n+1)} \][/tex]
We know:
[tex]\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \][/tex]
This series is telescoping, meaning most terms cancel out:
[tex]\[ \left(1 - \frac{1}{2} \right) + \left(\frac{1}{2} - \frac{1}{3} \right) + \left(\frac{1}{3} - \frac{1}{4} \right) + \ldots + \left(\frac{1}{2009} - \frac{1}{2010} \right) \][/tex]
After cancellation, we are left with:
[tex]\[ 1 - \frac{1}{2010} \][/tex]
So, the value of the expression is:
[tex]\[ 1 - \sum_{n=1}^{2009} \frac{1}{n(n+1)} = 1 - \left( 1 - \frac{1}{2010} \right) = \frac{1}{2010} \][/tex]
### Numerical Results:
1. The sum of the first 20 terms of the series [tex]\( 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \ldots \)[/tex] up to 20 terms is approximately 1.9047619047619044.
2. The value of [tex]\( 1 - \left[ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{2009 \times 2010} \right] \)[/tex] is approximately 0.0004975124378099327.
Thus, the numerical approach matches with the answer provided:
[tex]\[ (1.9047619047619044, 0.0004975124378099327) \][/tex]
