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, we need to find the value of \( a + b \) where:

- \( a_r = r \cdot {7 \choose r} \)
- \( b_r = (7 - r) \cdot {7 \choose r} \)
- \( A_r = \begin{pmatrix} a_r & 0 \\ 0 & b_r \end{pmatrix} \)

We are given that:

\[ A = \sum_{r=0}^{7} A_r = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \]

### Step 1: Calculate \( a \)

We start by calculating \( a \):

\[ a = \sum_{r=0}^{7} a_r = \sum_{r=0}^{7} r \cdot {7 \choose r} \]

Using the identity \( r \cdot {n \choose r} = n \cdot {n-1 \choose r-1} \), we can rewrite the summation:

\[ a = \sum_{r=1}^{7} 7 \cdot {6 \choose r-1} \]

This can be simplified further:

\[ a = 7 \sum_{r=1}^{7} {6 \choose r-1} \]

Changing the index of summation, let \( k = r - 1 \):

\[ a = 7 \sum_{k=0}^{6} {6 \choose k} = 7 \cdot 2^6 = 7 \cdot 64 = 448 \]

### Step 2: Calculate \( b \)

Next, we calculate \( b \):

\[ b = \sum_{r=0}^{7} b_r = \sum_{r=0}^{7} (7 - r) \cdot {7 \choose r} \]

This can be split into two parts:

\[ b = \sum_{r=0}^{7} 7 \cdot {7 \choose r} - \sum_{r=0}^{7} r \cdot {7 \choose r} \]

The first summation simplifies to:

\[ \sum_{r=0}^{7} 7 \cdot {7 \choose r} = 7 \cdot 2^7 = 7 \cdot 128 = 896 \]

We already calculated \( \sum_{r=0}^{7} r \cdot {7 \choose r} = a = 448 \).

Thus,

\[ b = 896 - 448 = 448 \]

### Step 3: Calculate \( a + b \)

Now we can find \( a + b \):

\[ a + b = 448 + 448 = 896 \]

### Final Answer

The value of \( a + b \) is:

\[ \boxed{896} \]