The sum of 10 terms of the series `1+2(1.1)+3(1.1)^(2)+4(1.1)^(3)+….` is

Options

A. 85.12
B. 92.5
C. 96.5
D. 100

Answer

To find the sum of the first 10 terms of the series \( S = 1 + 2(1.1) + 3(1.1)^2 + 4(1.1)^3 + \ldots + 10(1.1)^9 \), we can follow these steps:

### Step 1: Identify the series
The series can be expressed in a more manageable form:
\[ S = \sum_{n=1}^{10} n (1.1)^{n-1} \]
Here, \( x = 1.1 \).

### Step 2: Multiply the series by \( x \)
Multiply the entire series \( S \) by \( x \):
\[ xS = 1.1 \cdot 1 + 2(1.1) \cdot (1.1) + 3(1.1) \cdot (1.1)^2 + \ldots + 10(1.1) \cdot (1.1)^9 \]
This results in:
\[ xS = 1.1 + 2(1.1)^2 + 3(1.1)^3 + \ldots + 10(1.1)^{10} \]

### Step 3: Subtract the two series
Now, we subtract \( xS \) from \( S \):
\[ S - xS = (1 + 2(1.1) + 3(1.1)^2 + \ldots + 10(1.1)^9) - (1.1 + 2(1.1)^2 + 3(1.1)^3 + \ldots + 10(1.1)^{10}) \]
This simplifies to:
\[ S - xS = 1 + (2 - 1.1)(1.1) + (3 - 2)(1.1)^2 + (4 - 3)(1.1)^3 + \ldots + (10 - 9)(1.1)^9 - 10(1.1)^{10} \]
\[ = 1 + 0.9(1.1) + 1(1.1)^2 + 1(1.1)^3 + \ldots + 1(1.1)^9 - 10(1.1)^{10} \]

### Step 4: Recognize the geometric series
The remaining terms form a geometric series:
\[ = 1 + 0.9(1.1) + \sum_{k=2}^{9} (1.1)^k - 10(1.1)^{10} \]
The sum of the geometric series from \( k=0 \) to \( k=9 \) is:
\[ \sum_{k=0}^{9} (1.1)^k = \frac{(1.1)^{10} - 1}{1.1 - 1} = \frac{(1.1)^{10} - 1}{0.1} \]

### Step 5: Combine the results
Now we can express \( S \):
\[ S(1 - 1.1) = 1 + 0.9 \cdot \frac{(1.1)^{10} - 1}{0.1} - 10(1.1)^{10} \]
\[ S(-0.1) = 1 + 9(1.1)^{10} - 10(1.1)^{10} \]
\[ S(-0.1) = 1 - (1.1)^{10} \]
\[ S = \frac{1 - (1.1)^{10}}{-0.1} \]

### Step 6: Calculate \( (1.1)^{10} \)
Using a calculator or exponentiation:
\[ (1.1)^{10} \approx 2.59374 \]
Thus:
\[ S = \frac{1 - 2.59374}{-0.1} = \frac{-1.59374}{-0.1} \approx 15.9374 \]

### Final Answer
The sum of the first 10 terms of the series is approximately \( 15.9374 \).