Given `"^(8)C_(1)x(1-x)^(7)+2*^(8)C_(2)x^(2)(1-x)^(6)+3*^(8)C_(3)x^(3)(1-x)^(5)+...+8*x^(8)=ax+b`, then `a+b` is (a) 4 (b) 6 (c) 8 (d) 10

Options

A. `4`
B. `6`
C. `8`
D. `10`

Explanation

To solve the problem, we need to evaluate the expression given and find the values of \(a\) and \(b\) such that the expression can be represented as \(ax + b\).

The expression given is:

\[ \sum_{k=1}^{8} k \cdot \binom{8}{k} x^k (1-x)^{8-k} \]

This can be rewritten as:

\[ \binom{8}{1} x (1-x)^7 + 2 \cdot \binom{8}{2} x^2 (1-x)^6 + 3 \cdot \binom{8}{3} x^3 (1-x)^5 + \ldots + 8 \cdot x^8 \]

### Step 1: Recognizing the Structure
Notice that the powers of \(x\) and \(1-x\) always add up to 8. This is a key insight that will help us simplify the expression.

### Step 2: Evaluate at \(x = 1\)
To find \(a\) and \(b\), we can substitute \(x = 1\) into the expression.

When \(x = 1\):

- All terms with \(1-x\) will vanish (become 0) because they will have a positive power of \(1-x\).
- The only term that remains is the last term, which is \(8 \cdot \binom{8}{8} \cdot 1^8 = 8\).

Thus, we have:

\[ 8 = a \cdot 1 + b \]

This simplifies to:

\[ 8 = a + b \]

### Step 3: Finding \(a\) and \(b\)
Now, we need to find the coefficients \(a\) and \(b\).

To find \(a\), we can also evaluate the derivative of the original expression with respect to \(x\) and then substitute \(x = 1\).

The derivative of the expression gives us:

\[ \sum_{k=1}^{8} k \cdot \binom{8}{k} x^{k-1} (1-x)^{8-k} - \sum_{k=1}^{8} k \cdot \binom{8}{k} x^k (8-k)(1-x)^{7-k} \]

Evaluating this derivative at \(x = 1\) will yield the value of \(a\).

However, for simplicity, we can also observe that the coefficients of \(x\) in the original expression will yield \(a\) when evaluated at \(x = 1\).

### Step 4: Conclusion
From the above evaluations, we have:

\[ a + b = 8 \]

Thus, the value of \(a + b\) is:

\[ \boxed{8} \]