Find the `1+2.2+3.2^2+…….+t_n`

Options

A. `1+(1+n)2^n`
B. `1-(1+n)2^n`
C. `1-(1-n)2^n`
D. `1+(1-n)2^n`

Answer

To find the sum of the series \( S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + \ldots + n \cdot 2^{n-1} \), we can follow these steps:

### Step 1: Write the series in a more manageable form
We can express the series as:
\[ S = \sum_{k=1}^{n} k \cdot 2^{k-1} \]

### Step 2: Create a second series
To manipulate this series, we can multiply the entire series \( S \) by 2:
\[ 2S = \sum_{k=1}^{n} k \cdot 2^k \]

### Step 3: Align the two series
Now, we can write the two series:
1. \( S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + \ldots + n \cdot 2^{n-1} \)
2. \( 2S = 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + n \cdot 2^n \)

### Step 4: Subtract the two series
Now, subtract \( S \) from \( 2S \):
\[ 2S - S = S = (1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + n \cdot 2^n) - (1 + 2 \cdot 2 + 3 \cdot 2^2 + \ldots + n \cdot 2^{n-1}) \]
This simplifies to:
\[ S = (1 \cdot 2 - 1) + (2 \cdot 2^2 - 2 \cdot 2) + (3 \cdot 2^3 - 3 \cdot 2^2) + \ldots + (n \cdot 2^n - n \cdot 2^{n-1}) \]

### Step 5: Simplify the resulting series
This results in:
\[ S = 2 + (2^2) + (2^3) + \ldots + (2^n) - n \cdot 2^n \]
The first part is a geometric series:
\[ S = 2 + 2^2 + 2^3 + \ldots + 2^n - n \cdot 2^n \]

### Step 6: Calculate the sum of the geometric series
The sum of the geometric series \( 2 + 2^2 + 2^3 + \ldots + 2^n \) can be calculated using the formula for the sum of a geometric series:
\[ \text{Sum} = a \frac{r^n - 1}{r - 1} \]
where \( a = 2 \), \( r = 2 \), and \( n \) is the number of terms:
\[ \text{Sum} = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2 \]

### Step 7: Substitute back into the equation
Now substituting back into our equation for \( S \):
\[ S = (2^{n+1} - 2) - n \cdot 2^n \]
This simplifies to:
\[ S = 2^{n+1} - 2 - n \cdot 2^n \]

### Step 8: Final simplification
We can factor out \( 2^n \):
\[ S = 2^n(2 - n) - 2 \]

Thus, the final result for the sum of the series is:
\[ S = 2^n(2 - n) - 2 \]