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: Define the series
Let \( x = 1.1 \). Then the series can be rewritten as:
\[ S = 1 + 2x + 3x^2 + 4x^3 + \ldots + 10x^9 \]

### Step 2: Multiply the series by \( x \)
Now, multiply \( S \) by \( x \):
\[ xS = x + 2x^2 + 3x^3 + 4x^4 + \ldots + 10x^{10} \]

### Step 3: Subtract the two equations
Now, subtract the equation \( xS \) from \( S \):
\[ S - xS = (1 + 2x + 3x^2 + 4x^3 + \ldots + 10x^9) - (x + 2x^2 + 3x^3 + 4x^4 + \ldots + 10x^{10}) \]

This simplifies to:
\[ S - xS = 1 + (2x - x) + (3x^2 - 2x^2) + (4x^3 - 3x^3) + \ldots + (10x^9 - 9x^9) - 10x^{10} \]
\[ = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 - 10x^{10} \]

### Step 4: Factor out \( S \)
Factoring out \( S \):
\[ S(1 - x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 - 10x^{10} \]

### Step 5: Recognize the geometric series
The series \( 1 + x + x^2 + \ldots + x^9 \) is a geometric series with the first term \( a = 1 \) and common ratio \( r = x \). The sum of the first \( n \) terms of a geometric series is given by:
\[ \text{Sum} = \frac{a(r^n - 1)}{r - 1} \]
For our case:
\[ \text{Sum} = \frac{1(x^{10} - 1)}{x - 1} \]

### Step 6: Substitute back into the equation
Substituting this back, we have:
\[ S(1 - x) = \frac{x^{10} - 1}{x - 1} - 10x^{10} \]

### Step 7: Solve for \( S \)
Now, rearranging gives:
\[ S = \frac{\frac{x^{10} - 1}{x - 1} - 10x^{10}}{1 - x} \]

### Step 8: Substitute \( x = 1.1 \)
Now substituting \( x = 1.1 \):
\[ S = \frac{\frac{(1.1)^{10} - 1}{1.1 - 1} - 10(1.1)^{10}}{1 - 1.1} \]
Calculating \( (1.1)^{10} \):
\[ (1.1)^{10} \approx 2.59374 \]
Then substituting:
\[ S = \frac{\frac{2.59374 - 1}{0.1} - 10(2.59374)}{-0.1} \]
\[ = \frac{15.9374 - 25.9374}{-0.1} = \frac{-10}{-0.1} = 100 \]

### Final Answer
Thus, the sum of the first 10 terms of the series is:
\[ \boxed{100} \]