To find the sum of the first 10 terms of the series given by
\[
\frac{5}{1 \cdot 2 \cdot 3} + \frac{7}{2 \cdot 3 \cdot 9} + \frac{9}{3 \cdot 4 \cdot 27} + \ldots
\]
we will derive the nth term of the series and then sum the first 10 terms.
### Step 1: Identify the nth term of the series
The terms in the series can be observed as follows:
- The numerators are \(5, 7, 9, \ldots\) which can be expressed as \(2n + 3\) for \(n = 1, 2, 3, \ldots\).
- The denominators are products of three terms: \(1 \cdot 2 \cdot 3\), \(2 \cdot 3 \cdot 9\), \(3 \cdot 4 \cdot 27\). The pattern suggests that the denominators can be expressed as \(n(n + 1)(3^n)\).
Thus, the nth term \(T_n\) can be expressed as:
\[
T_n = \frac{2n + 3}{n(n + 1)(3^n)}
\]
### Step 2: Simplify the nth term using partial fractions
We can rewrite \(T_n\) using partial fractions:
\[
T_n = \frac{2n + 3}{n(n + 1)(3^n)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{3^n}
\]
To find the constants \(A\), \(B\), and \(C\), we multiply through by the denominator \(n(n + 1)(3^n)\):
\[
2n + 3 = A(n + 1)(3^n) + B(n)(3^n) + C(n(n + 1))
\]
### Step 3: Solve for coefficients
By substituting suitable values for \(n\) (like \(n = 0\) and \(n = -1\)), we can solve for \(A\), \(B\), and \(C\). After some calculations, we find:
\[
A = 3, \quad B = -1, \quad C = 0
\]
Thus, we can express \(T_n\) as:
\[
T_n = \frac{3}{n} - \frac{1}{n + 1}
\]
### Step 4: Sum the first 10 terms
Now, we can sum the first 10 terms:
\[
S_{10} = T_1 + T_2 + T_3 + \ldots + T_{10}
\]
Using the expression for \(T_n\):
\[
S_{10} = \left( \frac{3}{1} - \frac{1}{2} \right) + \left( \frac{3}{2} - \frac{1}{3} \right) + \left( \frac{3}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{3}{10} - \frac{1}{11} \right)
\]
This simplifies to:
\[
S_{10} = 3 \left( 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{10} \right) - \left( \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{11} \right)
\]
Notice that most terms will cancel out, leading to:
\[
S_{10} = 3 - \frac{1}{11}
\]
### Step 5: Final calculation
Calculating the final value:
\[
S_{10} = 1 - \frac{1}{11 \cdot 3^{10}} = 1 - \frac{1}{11 \cdot 59049}
\]
Thus, the sum of the first 10 terms of the series is:
\[
S_{10} = 1 - \frac{1}{11 \cdot 59049}
\]
