Find the sum of infinite terms of the series `3/1^2 + 5/(1^2 + 2^2) + 7/(1^2+2^2+3^2) + 9/(1^2+2^2+3^2+4^2)+...`

Answer

To find the sum of the infinite series given by:

\[ S = \frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \frac{9}{1^2 + 2^2 + 3^2 + 4^2} + \ldots \]

we will follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:

\[ T_n = \frac{2n + 1}{1^2 + 2^2 + 3^2 + \ldots + n^2} \]

### Step 2: Use the Formula for the Sum of Squares
The sum of the squares of the first \(n\) natural numbers is given by the formula:

\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Substituting this into our general term, we have:

\[ T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6(2n + 1)}{n(n + 1)(2n + 1)} \]

### Step 3: Simplify the General Term
Notice that \(2n + 1\) cancels out:

\[ T_n = \frac{6}{n(n + 1)} \]

### Step 4: Rewrite the General Term
We can further simplify \(T_n\) using partial fractions:

\[ T_n = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 5: Set Up the Infinite Series
Now, we can express the sum of the series:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 6: Recognize the Telescoping Series
The series is telescoping, meaning that most terms will cancel out:

\[ S = 6 \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots \right) \]

### Step 7: Evaluate the Limit
As \(n\) approaches infinity, the last term \(\frac{1}{n + 1}\) approaches 0. Therefore, we are left with:

\[ S = 6 \left( 1 - 0 \right) = 6 \]

### Final Answer
Thus, the sum of the infinite series is:

\[ \boxed{6} \]