Let `(1 + x + x^(2))^(100)=sum_(r=0)^(200)a_(r)x^(r)` and `a=sum_(r=0)^(200)a_(r)`, then value of `sum_(r=1)^(200)(ra_(r))/(25a)` is

Options

A. 5
B. 4
C. 3
D. 2

Answer

To solve the problem, we need to evaluate the expression \( \sum_{r=1}^{200} \frac{r a_r}{25a} \), where \( a = \sum_{r=0}^{200} a_r \) and \( (1 + x + x^2)^{100} = \sum_{r=0}^{200} a_r x^r \).

### Step 1: Find the value of \( a \)

We start by calculating \( a \):
\[ a = \sum_{r=0}^{200} a_r = (1 + 1 + 1^2)^{100} = 3^{100} \]

### Step 2: Find the value of \( \sum_{r=1}^{200} r a_r \)

Next, we need to find \( \sum_{r=1}^{200} r a_r \). We can find this by differentiating the generating function \( (1 + x + x^2)^{100} \):
\[ f(x) = (1 + x + x^2)^{100} \]
Differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = 100(1 + x + x^2)^{99}(1 + 2x) \]
Now, we evaluate \( f'(1) \):
\[ f'(1) = 100(1 + 1 + 1^2)^{99}(1 + 2) = 100 \cdot 3^{99} \cdot 3 = 300 \cdot 3^{99} \]
Thus,
\[ \sum_{r=1}^{200} r a_r = 300 \cdot 3^{99} \]

### Step 3: Substitute into the original expression

Now we substitute \( a \) and \( \sum_{r=1}^{200} r a_r \) into the expression:
\[ \sum_{r=1}^{200} \frac{r a_r}{25a} = \frac{1}{25a} \sum_{r=1}^{200} r a_r = \frac{1}{25 \cdot 3^{100}} \cdot (300 \cdot 3^{99}) \]
This simplifies to:
\[ = \frac{300 \cdot 3^{99}}{25 \cdot 3^{100}} = \frac{300}{25 \cdot 3} = \frac{300}{75} = 4 \]

### Final Answer

Thus, the value of \( \sum_{r=1}^{200} \frac{r a_r}{25a} \) is \( \boxed{4} \).