Find the sum `(1xx2)/(3!)+(2xx(2)^2)/(4!)+(3xx(2)^3)/(5!)+ . . . +(20xx(2)^20)/(22!)`

Explanation

To find the sum

\[ S = \frac{1 \times 2}{3!} + \frac{2 \times (2^2)}{4!} + \frac{3 \times (2^3)}{5!} + \ldots + \frac{20 \times (2^{20})}{22!}, \]

we start by identifying the general term of the series. The \( n \)-th term can be expressed as:

\[ T_n = \frac{n \times (2^n)}{(n+2)!}. \]

### Step 1: Rewrite the general term

We can rewrite the general term \( T_n \) as follows:

\[ T_n = \frac{n \times 2^n}{(n+2)!} = \frac{n \times 2^n}{(n+2)(n+1)n!} = \frac{2^n}{(n+1)!} - \frac{2^n}{(n+2)!}. \]

### Step 2: Sum the series

Now, we can express the sum \( S \) as:

\[ S = \sum_{n=1}^{20} \left( \frac{2^n}{(n+1)!} - \frac{2^n}{(n+2)!} \right). \]

### Step 3: Break the summation into two parts

We can separate the summation:

\[ S = \sum_{n=1}^{20} \frac{2^n}{(n+1)!} - \sum_{n=1}^{20} \frac{2^n}{(n+2)!}. \]

### Step 4: Change the index of summation

For the second summation, we can change the index by letting \( m = n + 1 \):

\[ \sum_{n=1}^{20} \frac{2^n}{(n+2)!} = \sum_{m=2}^{21} \frac{2^{m-1}}{m!}. \]

### Step 5: Combine the two sums

Now we have:

\[ S = \sum_{n=1}^{20} \frac{2^n}{(n+1)!} - \sum_{m=2}^{21} \frac{2^{m-1}}{m!}. \]

### Step 6: Evaluate the sums

Now we can evaluate the sums:

1. The first sum is:

\[ \sum_{n=1}^{20} \frac{2^n}{(n+1)!} = \sum_{k=2}^{21} \frac{2^{k-1}}{k!} = \frac{1}{2} \sum_{k=2}^{21} \frac{2^k}{k!}. \]

2. The second sum can be expressed as:

\[ \sum_{m=2}^{21} \frac{2^{m-1}}{m!} = \frac{1}{2} \sum_{k=2}^{21} \frac{2^k}{k!}. \]

### Step 7: Final simplification

Thus, we can see that:

\[ S = \frac{1}{2} \sum_{k=2}^{21} \frac{2^k}{k!} - \frac{1}{2} \sum_{k=2}^{21} \frac{2^k}{k!} = \frac{1}{2} \left( \frac{2^2}{2!} - \frac{2^{21}}{22!} \right). \]

### Step 8: Calculate the final value

Now we can compute the final value:

\[ S = \frac{2}{2} - \frac{2^{21}}{22!} = 1 - \frac{2^{21}}{22!}. \]

Thus, the final answer is:

\[ S = 1 - \frac{2^{21}}{22!}. \]