For r = 0,1,...10, let `A_(r ),B_(r )` and `C_(r )` denote respectively the coefficient of `x^(r )` in the expansions of `(1+x)^(10), (1+x)^(20)` and `(1+x)^(30)`. Then `sum_(r=1)^(10) A_(r ) (B_(10)B_(r )-C_(10)A_(r ))` is equal to

Options

A. `B_(10)-C_(10)`
B. `A_(10) (B_(10)^(2) -C_(10)A_(10))`
C. 0
D. `C_(10)-B_(10)`

Answer

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

\[ \sum_{r=1}^{10} A_r (B_{10} B_r - C_{10} A_r) \]

where \( A_r \), \( B_r \), and \( C_r \) are the coefficients of \( x^r \) in the expansions of \( (1+x)^{10} \), \( (1+x)^{20} \), and \( (1+x)^{30} \) respectively.

### Step 1: Identify the coefficients
The coefficients can be expressed using binomial coefficients:
- \( A_r = \binom{10}{r} \)
- \( B_r = \binom{20}{r} \)
- \( C_r = \binom{30}{r} \)

### Step 2: Substitute the coefficients into the sum
Substituting these coefficients into the sum gives:

\[ \sum_{r=1}^{10} \binom{10}{r} \left( \binom{20}{10} \binom{20}{r} - \binom{30}{10} \binom{10}{r} \right) \]

### Step 3: Distribute the summation
Distributing the summation yields two separate sums:

\[ \sum_{r=1}^{10} \binom{10}{r} \binom{20}{10} \binom{20}{r} - \sum_{r=1}^{10} \binom{10}{r} \binom{30}{10} \binom{10}{r} \]

### Step 4: Factor out constants
The first term can be simplified as:

\[ \binom{20}{10} \sum_{r=1}^{10} \binom{10}{r} \binom{20}{r} \]

The second term can be simplified as:

\[ \binom{30}{10} \sum_{r=1}^{10} \binom{10}{r} \binom{10}{r} \]

### Step 5: Evaluate the sums
Using the binomial theorem, we know that:

\[ \sum_{r=0}^{n} \binom{n}{r} \binom{m}{r} = \binom{n+m}{n} \]

For the first sum:

\[ \sum_{r=0}^{10} \binom{10}{r} \binom{20}{r} = \binom{30}{10} \]

Thus, we can write:

\[ \sum_{r=1}^{10} \binom{10}{r} \binom{20}{r} = \binom{30}{10} - \binom{10}{0} \binom{20}{0} = \binom{30}{10} - 1 \]

For the second sum:

\[ \sum_{r=0}^{10} \binom{10}{r}^2 = \binom{20}{10} \]

Thus, we can write:

\[ \sum_{r=1}^{10} \binom{10}{r}^2 = \binom{20}{10} - 1 \]

### Step 6: Substitute back into the expression
Now substituting these results back into our expression gives:

\[ \binom{20}{10} \left( \binom{30}{10} - 1 \right) - \binom{30}{10} \left( \binom{20}{10} - 1 \right) \]

### Step 7: Simplify the expression
This simplifies to:

\[ \binom{20}{10} \binom{30}{10} - \binom{20}{10} - \binom{30}{10} \binom{20}{10} + \binom{30}{10} \]

The terms \( \binom{20}{10} \binom{30}{10} \) cancel out:

\[ \binom{30}{10} - \binom{20}{10} \]

### Final Result
Thus, the final result is:

\[ \boxed{\binom{30}{10} - \binom{20}{10}} \]