Find sum to infinite terms of the series `1 + 2 ((11)/(10)) + 3 ((11)/(10)) ^(2) + 4 ((11)/(10)) ^(3) +………`

Answer

To find the sum to infinite terms of the series \( S = 1 + 2 \left(\frac{11}{10}\right) + 3 \left(\frac{11}{10}\right)^2 + 4 \left(\frac{11}{10}\right)^3 + \ldots \), we can follow these steps:

### Step 1: Identify the series
The series can be expressed as:
\[ S = \sum_{n=1}^{\infty} n \left(\frac{11}{10}\right)^{n-1} \]
This is a series where the coefficients \( n \) form an arithmetic progression and the terms \( \left(\frac{11}{10}\right)^{n-1} \) form a geometric progression.

### Step 2: Use the formula for the sum of an infinite series
The sum of an infinite series of the form \( S = \sum_{n=1}^{\infty} n x^{n-1} \) can be calculated using the formula:
\[ S = \frac{1}{(1-x)^2} \]
where \( |x| < 1 \). Here, we have \( x = \frac{11}{10} \), which is greater than 1. However, we will manipulate the series to find the sum.

### Step 3: Multiply the series by \( \frac{11}{10} \)
Let’s multiply the entire series \( S \) by \( \frac{11}{10} \):
\[ \frac{11}{10} S = \frac{11}{10} \left( 1 + 2 \left(\frac{11}{10}\right) + 3 \left(\frac{11}{10}\right)^2 + 4 \left(\frac{11}{10}\right)^3 + \ldots \right) \]
This gives us:
\[ \frac{11}{10} S = \frac{11}{10} + 2 \left(\frac{11}{10}\right)^2 + 3 \left(\frac{11}{10}\right)^3 + 4 \left(\frac{11}{10}\right)^4 + \ldots \]

### Step 4: Set up the equation
Now, we can set up the equation:
\[ S - \frac{11}{10} S = 1 + \left(2 - \frac{11}{10}\right) \left(\frac{11}{10}\right) + \left(3 - 2\right) \left(\frac{11}{10}\right)^2 + \left(4 - 3\right) \left(\frac{11}{10}\right)^3 + \ldots \]
This simplifies to:
\[ S - \frac{11}{10} S = 1 + \left(\frac{1}{10}\right) \left(\frac{11}{10}\right) + \left(1\right) \left(\frac{11}{10}\right)^2 + \left(1\right) \left(\frac{11}{10}\right)^3 + \ldots \]

### Step 5: Recognize the remaining series
The remaining series is a geometric series:
\[ \sum_{n=0}^{\infty} \left(\frac{11}{10}\right)^n = \frac{1}{1 - \frac{11}{10}} = -10 \]
Thus, we have:
\[ S - \frac{11}{10} S = 1 - 10 \]
\[ S - \frac{11}{10} S = -9 \]

### Step 6: Solve for \( S \)
Now, factor out \( S \):
\[ S \left(1 - \frac{11}{10}\right) = -9 \]
\[ S \left(-\frac{1}{10}\right) = -9 \]
\[ S = -9 \cdot -10 = 90 \]

### Final Answer
The sum to infinite terms of the series is:
\[ \boxed{90} \]