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. (a) `3`
B. (b) `5`
C. (c) `13`
D. (d) `2^(41)`

Explanation

To solve the problem, we need to find the sum

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

### Step 1: Identify the general term

The coefficients in front of the binomial coefficients can be expressed in terms of \( R \):

- For \( R = 2 \), the coefficient is \( 2 = 2^2 - 2 \)
- For \( R = 3 \), the coefficient is \( 6 = 3^2 - 3 \)
- For \( R = 4 \), the coefficient is \( 12 = 4^2 - 4 \)
- For \( R = 5 \), the coefficient is \( 20 = 5^2 - 5 \)
- ...
- For \( R = 40 \), the coefficient is \( 1560 = 40^2 - 40 \)

Thus, the general term can be expressed as:

\[ T_R = (R^2 - R) \cdot \binom{40}{R} \]

### Step 2: Rewrite the general term

We can factor \( R \) out of the expression:

\[ T_R = R(R - 1) \cdot \binom{40}{R} \]

Using the identity \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), we can express \( \binom{40}{R} \) as:

\[ \binom{40}{R} = \frac{40!}{R!(40-R)!} \]

### Step 3: Simplify the sum

Now, we can rewrite the sum \( S \):

\[ S = \sum_{R=2}^{40} R(R-1) \cdot \binom{40}{R} \]

This can be simplified using the identity:

\[ R(R-1) \cdot \binom{n}{R} = n(n-1) \cdot \binom{n-2}{R-2} \]

Thus, we can express \( S \) as:

\[ S = 40 \cdot 39 \cdot \sum_{R=2}^{40} \binom{38}{R-2} \]

### Step 4: Change the index of summation

Changing the index of summation by letting \( k = R - 2 \):

\[ S = 40 \cdot 39 \cdot \sum_{k=0}^{38} \binom{38}{k} \]

### Step 5: Evaluate the sum of binomial coefficients

Using the binomial theorem, we know:

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

Thus,

\[ \sum_{k=0}^{38} \binom{38}{k} = 2^{38} \]

### Step 6: Substitute back into the expression for \( S \)

Substituting this back into our expression for \( S \):

\[ S = 40 \cdot 39 \cdot 2^{38} \]

### Step 7: Factor the expression

Now, we can factor \( S \):

\[ S = 5 \cdot 8 \cdot 39 \cdot 2^{38} \]

### Step 8: Factor \( 39 \)

The number \( 39 \) can be factored as:

\[ 39 = 3 \cdot 13 \]

Thus, we have:

\[ S = 5 \cdot 3 \cdot 13 \cdot 8 \cdot 2^{38} \]

### Conclusion

The final expression shows that \( S \) is divisible by \( 5, 3, 13, \) and \( 2^{41} \) (since \( 8 = 2^3 \) and \( 2^{38} \) gives \( 2^{41} \) when combined).