The sum of the series `2.""^(20)C_0+5.""^(20)C_1+8.""^(20)C_2+11.""^(20)C_(3)+.........62.^(20)C_20` is equal

Options

A. `2^(24)`
B. `2^(25)`
C. `2^(26)`
D. `2^(23)`

Answer

To solve the given series \( S = 2 \cdot \binom{20}{0} + 5 \cdot \binom{20}{1} + 8 \cdot \binom{20}{2} + 11 \cdot \binom{20}{3} + \ldots + 62 \cdot \binom{20}{20} \), we can follow these steps:

### Step 1: Identify the Pattern
The coefficients in the series are \( 2, 5, 8, 11, \ldots, 62 \). We can observe that these coefficients form an arithmetic sequence where the first term \( a = 2 \) and the common difference \( d = 3 \).

### Step 2: General Term of the Coefficients
The \( n \)-th term of the arithmetic sequence can be expressed as:
\[ a_n = 2 + (n-1) \cdot 3 = 3n - 1 \]
Thus, we can rewrite the series as:
\[ S = \sum_{n=0}^{20} (3n - 1) \cdot \binom{20}{n} \]

### Step 3: Split the Series
We can split the sum into two separate sums:
\[ S = \sum_{n=0}^{20} (3n) \cdot \binom{20}{n} - \sum_{n=0}^{20} \binom{20}{n} \]

### Step 4: Evaluate the Second Sum
The second sum, \( \sum_{n=0}^{20} \binom{20}{n} \), is known to equal \( 2^{20} \) (by the binomial theorem).

### Step 5: Evaluate the First Sum
For the first sum, we can use the identity \( n \cdot \binom{n}{k} = k \cdot \binom{n-1}{k-1} \):
\[ \sum_{n=0}^{20} n \cdot \binom{20}{n} = 20 \cdot 2^{19} \]
Thus,
\[ \sum_{n=0}^{20} 3n \cdot \binom{20}{n} = 3 \cdot 20 \cdot 2^{19} = 60 \cdot 2^{19} \]

### Step 6: Combine the Results
Now we can substitute back into our expression for \( S \):
\[ S = 60 \cdot 2^{19} - 2^{20} \]

### Step 7: Simplify
We can factor out \( 2^{19} \):
\[ S = 2^{19} (60 - 2) = 2^{19} \cdot 58 \]

### Step 8: Final Expression
Now, we can express \( 58 \) as \( 2 \cdot 29 \):
\[ S = 2^{20} \cdot 29 \]

### Conclusion
Thus, the sum of the series is:
\[ S = 2^{20} \cdot 29 \]