Series

This series seems to be a telescoping series where each term is of the form \(\frac{1}{(3n-2)(3n+1)}\). We can decompose each fraction into partial fractions to simplify the series.

 

Let's rewrite each term:

 

\[\frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1}\]

 

Now, let's find the values of \(A\) and \(B\). We'll multiply both sides by \((3n-2)(3n+1)\):

 

\[1 = A(3n+1) + B(3n-2)\]

 

Let \(n = \frac{2}{3}\), then we get:

 

\[1 = A + 0 \Rightarrow A = 1\]

 

Let \(n = -1\), then we get:

 

\[1 = 0 + 3B \Rightarrow B = \frac{1}{3}\]

 

So, the decomposition becomes:

 

\[\frac{1}{(3n-2)(3n+1)} = \frac{1}{3n-2} + \frac{1}{3(3n+1)}\]

 

Now, let's rewrite the series using these decompositions:

 

\[\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \dots + \frac{1}{37 \times 40}\]

 

\[= \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \dots + \left(\frac{1}{37} - \frac{1}{40}\right)\]

 

Now, observe how most of the terms cancel out:

 

\[= \frac{1}{1} - \frac{1}{40}\]

 

\[= 1 - \frac{1}{40}\]

 

\[= \frac{39}{40}\]

 

So, the sum of the series is \(\frac{39}{40}\).