let `2..^(20)C_(0)+5.^(20)C_(1)+8.^(20)C_(2)+?.+62.^(20)C_(20)`. Then sum of this series is

Options

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

Answer

To solve the problem, we need to evaluate the sum:

\[ S = 2 \cdot \binom{20}{0} + 5 \cdot \binom{20}{1} + 8 \cdot \binom{20}{2} + \ldots + 62 \cdot \binom{20}{20} \]

### Step 1: Express the series in a more manageable form
We can rewrite the series as:

\[ S = \sum_{k=0}^{20} (2 + 3k) \cdot \binom{20}{k} \]

This is because the coefficients \(2, 5, 8, \ldots, 62\) can be expressed as \(2 + 3k\) where \(k\) ranges from \(0\) to \(20\).

### Step 2: Split the sum
We can split \(S\) into two separate sums:

\[ S = \sum_{k=0}^{20} 2 \cdot \binom{20}{k} + \sum_{k=0}^{20} 3k \cdot \binom{20}{k} \]

### Step 3: Evaluate the first sum
The first sum is:

\[ \sum_{k=0}^{20} 2 \cdot \binom{20}{k} = 2 \cdot \sum_{k=0}^{20} \binom{20}{k} = 2 \cdot 2^{20} \]

This is because the sum of binomial coefficients \(\sum_{k=0}^{n} \binom{n}{k} = 2^n\).

### Step 4: Evaluate the second sum
The second sum can be evaluated using the identity \(k \cdot \binom{n}{k} = n \cdot \binom{n-1}{k-1}\):

\[ \sum_{k=0}^{20} 3k \cdot \binom{20}{k} = 3 \cdot 20 \cdot \sum_{k=1}^{20} \binom{19}{k-1} = 3 \cdot 20 \cdot 2^{19} \]

### Step 5: Combine the results
Now we can combine both sums:

\[ S = 2 \cdot 2^{20} + 60 \cdot 2^{19} \]

Factoring out \(2^{19}\):

\[ S = 2^{19} (2 \cdot 2 + 60) = 2^{19} (4 + 60) = 2^{19} \cdot 64 \]

### Step 6: Simplify the expression
Now we simplify:

\[ S = 64 \cdot 2^{19} = 2^6 \cdot 2^{19} = 2^{25} \]

### Final Answer
Thus, the sum of the series is:

\[ \boxed{2^{25}} \]