+0

# Series

+1
6
1
+246

Compute
$$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \dots + \frac{1}{37 \times 40}.$$

Apr 16, 2024

#1
+415
0

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 $$\frac{39}{40}$$.

Apr 28, 2024