Let `a_r=r(.^7C_r),b_r=(7-r)(.^7C_r)` and `A_r=[(a_r,0),(0,b_r)]`. If `A=sum_(r=0)^7 A_r=[(a,0),(0,b)]`, then find the value of a+b .

Answer

To solve the problem step by step, we will first define the terms and then compute the required sums.

### Step 1: Define the terms
We have:
- \( a_r = r \cdot \binom{7}{r} \)
- \( b_r = (7 - r) \cdot \binom{7}{r} \)

### Step 2: Write the matrices
The matrices \( A_r \) are given as:
\[ A_r = \begin{pmatrix} a_r & 0 \\ 0 & b_r \end{pmatrix} \]
We need to find the sum of these matrices from \( r = 0 \) to \( r = 7 \):
\[ A = \sum_{r=0}^{7} A_r = \sum_{r=0}^{7} \begin{pmatrix} a_r & 0 \\ 0 & b_r \end{pmatrix} = \begin{pmatrix} \sum_{r=0}^{7} a_r & 0 \\ 0 & \sum_{r=0}^{7} b_r \end{pmatrix} \]

### Step 3: Calculate \( a \)
Now, we calculate \( a \):
\[ a = \sum_{r=0}^{7} a_r = \sum_{r=0}^{7} r \cdot \binom{7}{r} \]
Using the identity \( r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \), we can rewrite this as:
\[ a = 7 \cdot \sum_{r=1}^{7} \binom{6}{r-1} = 7 \cdot \sum_{s=0}^{6} \binom{6}{s} = 7 \cdot 2^6 = 7 \cdot 64 = 448 \]

### Step 4: Calculate \( b \)
Next, we calculate \( b \):
\[ b = \sum_{r=0}^{7} b_r = \sum_{r=0}^{7} (7 - r) \cdot \binom{7}{r} \]
This can be rewritten as:
\[ b = 7 \cdot \sum_{r=0}^{7} \binom{7}{r} - \sum_{r=0}^{7} r \cdot \binom{7}{r} \]
We know that \( \sum_{r=0}^{7} \binom{7}{r} = 2^7 = 128 \) and we already calculated \( \sum_{r=0}^{7} r \cdot \binom{7}{r} = 448 \). Thus,
\[ b = 7 \cdot 128 - 448 = 896 - 448 = 448 \]

### Step 5: Calculate \( a + b \)
Finally, we find:
\[ a + b = 448 + 448 = 896 \]

### Final Answer
The value of \( a + b \) is \( \boxed{896} \).