If `(1+x^(2))^(n) =sum_(r=0)^(n) a_(r)x^(r )= (1+x+x^(2)+x^(3))^(100)`. If `a = sum_(r=0)^(300)a_(r)`, then `sum_(r=0)^(300) ra_(r)` is

Explanation

To solve the problem, we need to find the value of \( \sum_{r=0}^{300} r a_r \) given that \( (1+x^2)^n = \sum_{r=0}^{n} a_r x^r = (1+x+x^2+x^3)^{100} \).

### Step 1: Understanding the Given Expression
We have:
\[ (1+x^2)^n = (1+x+x^2+x^3)^{100} \]
This means that the coefficients \( a_r \) in the expansion of \( (1+x^2)^n \) are equal to the coefficients of \( x^r \) in the expansion of \( (1+x+x^2+x^3)^{100} \).

### Step 2: Finding \( a \)
We need to find:
\[ a = \sum_{r=0}^{300} a_r \]
To find \( a \), we can substitute \( x = 1 \) in the expression:
\[ (1 + 1 + 1 + 1)^{100} = 4^{100} \]
Thus, we have:
\[ a = 4^{100} \]

### Step 3: Finding \( \sum_{r=0}^{300} r a_r \)
We want to find:
\[ \sum_{r=0}^{300} r a_r \]
To do this, we differentiate the generating function:
\[ \sum_{r=0}^{300} a_r x^r = (1+x+x^2+x^3)^{100} \]
Differentiating both sides with respect to \( x \):
\[ \sum_{r=0}^{300} r a_r x^{r-1} = 100(1+x+x^2+x^3)^{99}(1 + 2x + 3x^2) \]
Now, we substitute \( x = 1 \):
\[ \sum_{r=0}^{300} r a_r = 100(1+1+1+1)^{99}(1 + 2 + 3) = 100 \cdot 4^{99} \cdot 6 \]

### Step 4: Simplifying the Expression
We can simplify this:
\[ \sum_{r=0}^{300} r a_r = 600 \cdot 4^{99} \]

### Step 5: Expressing in Terms of \( a \)
Since \( a = 4^{100} \), we can express \( 4^{99} \) as:
\[ 4^{99} = \frac{a}{4} \]
Thus:
\[ \sum_{r=0}^{300} r a_r = 600 \cdot \frac{a}{4} = 150a \]

### Final Answer
The final result is:
\[ \sum_{r=0}^{300} r a_r = 150a \]