To solve the given series, we'll sum the terms of the series:
[tex]\[ \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{5 \cdot 6 \cdot 7} + \ldots + \frac{1}{23 \cdot 24 \cdot 25} \][/tex]
### Step-by-Step Solution:
1. Identify the General Term:
We have to find the general expression for the series. The [tex]\(i\)[/tex]-th term in the series can be written as:
[tex]\[
\frac{1}{i(i+1)(i+2)}
\][/tex]
Here, [tex]\(i\)[/tex] starts from 2 and goes up to 23.
2. Expand the Expression:
Using partial fraction decomposition, we can express the general term [tex]\(\frac{1}{i(i+1)(i+2)}\)[/tex] in a simpler form:
[tex]\[
\frac{1}{i(i+1)(i+2)} = \frac{A}{i} + \frac{B}{i+1} + \frac{C}{i+2}
\][/tex]
3. Find Constants [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
To find [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex], we'll solve the equation:
[tex]\[
1 = A(i+1)(i+2) + B(i)(i+2) + C(i)(i+1)
\][/tex]
By substituting appropriate values of [tex]\(i\)[/tex], we can determine [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex].
After solving, we get:
[tex]\[
A = \frac{1}{2},\quad B = -1, \quad \text{and} \quad C = \frac{1}{2}
\][/tex]
Thus,
[tex]\[
\frac{1}{i(i+1)(i+2)} = \frac{1}{2i} - \frac{1}{i+1} + \frac{1}{2(i+2)}
\][/tex]
4. Sum the Series:
[tex]\[
\sum_{i=2}^{23} \left( \frac{1}{2i} - \frac{1}{i+1} + \frac{1}{2(i+2)} \right)
\][/tex]
5. Separate and Simplify the Series:
The series can be broken down into sums of three simpler series:
[tex]\[
\sum_{i=2}^{23} \frac{1}{2i} - \sum_{i=2}^{23} \frac{1}{i+1} + \sum_{i=2}^{23} \frac{1}{2(i+2)}
\][/tex]
Each of these series can be evaluated by recognizing they are telescoping series.
6. Calculate the Series:
Since the terms in the series are fractions of products, it simplifies to adding up the fractions:
The sum of the given series, after adding each term from [tex]\(i = 2\)[/tex] to [tex]\(i = 23\)[/tex], is:
[tex]\[
\boxed{0.0825}
\][/tex]
Thus, the sum of the series [tex]\(\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\frac{1}{4 \cdot 5 \cdot 6}+\frac{1}{5 \cdot 6 \cdot 7}+\ldots .+\frac{1}{23 \cdot 24 \cdot 25}\)[/tex] is approximately 0.0825.
