The value of `sum_(r=0)^(40) r""^(40)C_(r)""^(30)C_(r)` is (a) `40.^(69)C_(29)` (b) `40.^(70)C_(30)` (c) `.^(60)C_(29)` (d) `.^(70)C_(30)`

Options

A. `40.^(69)C_(29)`
B. `40.^(70)C_(30)`
C. `.^(60)C_(29)`
D. `.^(70)C_(30)`

Explanation

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

\[ \sum_{r=0}^{40} r \cdot \binom{40}{r} \cdot \binom{30}{r} \]

### Step 1: Rewrite the term \( r \cdot \binom{40}{r} \)

We can express \( r \cdot \binom{40}{r} \) in terms of binomial coefficients. Recall that:

\[ r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \]

Applying this to our case:

\[ r \cdot \binom{40}{r} = 40 \cdot \binom{39}{r-1} \]

### Step 2: Substitute into the sum

Now we substitute this back into our summation:

\[ \sum_{r=0}^{40} r \cdot \binom{40}{r} \cdot \binom{30}{r} = \sum_{r=0}^{40} 40 \cdot \binom{39}{r-1} \cdot \binom{30}{r} \]

### Step 3: Change the index of summation

To simplify the summation, we change the index of summation by letting \( k = r - 1 \). Then when \( r = 0 \), \( k = -1 \) (which we can ignore since \( \binom{39}{-1} = 0 \)), and when \( r = 40 \), \( k = 39 \). Thus, we can rewrite the sum as:

\[ 40 \cdot \sum_{k=0}^{39} \binom{39}{k} \cdot \binom{30}{k+1} \]

### Step 4: Use the Vandermonde Identity

The sum \( \sum_{k=0}^{n} \binom{m}{k} \cdot \binom{n}{r-k} = \binom{m+n}{r} \) can be applied here. In our case, we have:

\[ \sum_{k=0}^{39} \binom{39}{k} \cdot \binom{30}{k+1} = \sum_{k=0}^{39} \binom{39}{k} \cdot \binom{30}{30-(k+1)} = \sum_{k=0}^{39} \binom{39}{k} \cdot \binom{30}{29-k} \]

This is equivalent to:

\[ \binom{39 + 30}{30} = \binom{69}{30} \]

### Step 5: Final expression

Now substituting this back into our expression, we have:

\[ 40 \cdot \binom{69}{30} \]

### Conclusion

Thus, the value of the original sum is:

\[ \sum_{r=0}^{40} r \cdot \binom{40}{r} \cdot \binom{30}{r} = 40 \cdot \binom{69}{30} \]

The correct answer is:

(b) \( 40 \cdot \binom{70}{30} \)