The sum `2 xx .^(40)C_(2) + 6 xx .^(40)C_(3) + 12 xx .^(40)C_(4) + 20 xx .^(40)C_(5) + "…." + 1560 xx .^(40)C_(40)` is divisible by

Options

A. `3`
B. `5`
C. `13`
D. `2^(41)`

Explanation

To solve the problem, we need to analyze the given sum:

\[ S = 2 \cdot \binom{40}{2} + 6 \cdot \binom{40}{3} + 12 \cdot \binom{40}{4} + 20 \cdot \binom{40}{5} + \ldots + 1560 \cdot \binom{40}{40} \]

We can observe that the coefficients \(2, 6, 12, 20, \ldots, 1560\) can be expressed in terms of \(r\), where \(r\) is the index of the binomial coefficient.

### Step 1: Identify the pattern in the coefficients
The coefficients can be represented as \(r(r-1)\) for \(r = 2, 3, 4, \ldots, 40\). This is because:
- For \(r = 2\), the coefficient is \(2(2-1) = 2\)
- For \(r = 3\), the coefficient is \(3(3-1) = 6\)
- For \(r = 4\), the coefficient is \(4(4-1) = 12\)
- For \(r = 5\), the coefficient is \(5(5-1) = 20\)
- ...
- For \(r = 40\), the coefficient is \(40(40-1) = 1560\)

### Step 2: Rewrite the sum
Using the identified pattern, we can rewrite the sum as:
\[ S = \sum_{r=2}^{40} r(r-1) \cdot \binom{40}{r} \]

### Step 3: Factor out constants
We can factor \(r(r-1)\) as:
\[ r(r-1) = 40 \cdot \binom{39}{r-2} \]
Thus, we can express the sum as:
\[ S = 40 \cdot \sum_{r=2}^{40} \binom{39}{r-2} \]

### Step 4: Change the index of summation
Let \(k = r - 2\). Then when \(r = 2\), \(k = 0\) and when \(r = 40\), \(k = 38\). Therefore, we can rewrite the sum:
\[ S = 40 \cdot \sum_{k=0}^{38} \binom{39}{k} \]

### Step 5: Evaluate the summation
Using the binomial theorem, we know that:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
Thus:
\[ \sum_{k=0}^{38} \binom{39}{k} = 2^{39} \]

### Step 6: Final expression for S
Substituting back, we get:
\[ S = 40 \cdot 2^{39} \]

### Step 7: Determine divisibility
Now, we need to find what \(S\) is divisible by:
\[ S = 40 \cdot 2^{39} = 2^3 \cdot 5 \cdot 2^{39} = 2^{42} \cdot 5 \]

### Conclusion
The sum \(S\) is divisible by \(2^{42}\) and \(5\). Therefore, the answer is that \(S\) is divisible by \(2^{42}\) and \(5\).