Sum up to n terms the series
`2.1+3.2+4.4+5.8+.....`

Answer

To find the sum up to n terms of the series \(2 \cdot 1 + 3 \cdot 2 + 4 \cdot 4 + 5 \cdot 8 + \ldots\), we can follow these steps:

### Step 1: Identify the pattern in the series
The series can be expressed as:
\[ S_n = 2 \cdot 1 + 3 \cdot 2 + 4 \cdot 4 + 5 \cdot 8 + \ldots \]
We can observe that the coefficients of the terms are \(2, 3, 4, 5, \ldots\) which form an arithmetic progression (AP) starting from 2 with a common difference of 1. The second part of each term \(1, 2, 4, 8, \ldots\) forms a geometric progression (GP) with a first term of 1 and a common ratio of 2.

### Step 2: Write the nth term
The nth term of the series can be expressed as:
\[ T_n = (n + 1) \cdot 2^{n-1} \]
where \(n + 1\) comes from the arithmetic progression (starting from 2) and \(2^{n-1}\) comes from the geometric progression.

### Step 3: Write the sum of the first n terms
The sum of the first n terms \(S_n\) can be expressed as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k + 1) \cdot 2^{k-1} \]

### Step 4: Split the sum
We can split the sum into two parts:
\[ S_n = \sum_{k=1}^{n} k \cdot 2^{k-1} + \sum_{k=1}^{n} 2^{k-1} \]

### Step 5: Calculate the second sum
The second sum is a geometric series:
\[ \sum_{k=1}^{n} 2^{k-1} = 1 + 2 + 4 + \ldots + 2^{n-1} = 2^n - 1 \]

### Step 6: Calculate the first sum
For the first sum, we can use the formula for the sum of \(k \cdot r^{k-1}\):
\[ \sum_{k=1}^{n} k \cdot r^{k-1} = r \frac{d}{dr} \left( \sum_{k=0}^{n} r^k \right) \]
The sum of the geometric series is:
\[ \sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r} \]
Differentiating with respect to \(r\) and multiplying by \(r\) gives us:
\[ \sum_{k=1}^{n} k \cdot r^{k-1} = \frac{r(1 - (n+1)r^n + nr^{n+1})}{(1 - r)^2} \]
Substituting \(r = 2\):
\[ \sum_{k=1}^{n} k \cdot 2^{k-1} = \frac{2(1 - (n+1)2^n + n \cdot 2^{n+1})}{(1 - 2)^2} = 2(1 - (n+1)2^n + n \cdot 2^{n+1}) \]

### Step 7: Combine the results
Now we combine both sums:
\[ S_n = 2(1 - (n+1)2^n + n \cdot 2^{n+1}) + (2^n - 1) \]
Simplifying this expression will give us the final result for \(S_n\).

### Final Result
After simplification, we find:
\[ S_n = 2^{n+1}(n - 1) + 2 \]