Let `S = sum _(r=1)^(30) (""^(30+r)C_(r) (2r-1))/(""^(30)C_(r)(30+r)),K=sum_(r=0)^(30) (""^(30)C_(r))^(2)`
and `G=sum_(r=0)^(60) (-1)^(r)(""^(60)C_(r) )^(2)`
The value of K + G is

Options

A. 2 S - 2
B. 2 S - 1
C. 2 S + 1
D. 2 S + 2

Explanation

To solve the problem, we need to evaluate the expressions for \( S \), \( K \), and \( G \) and then find the value of \( K + G \).

### Step-by-Step Solution:

1. **Evaluate \( S \)**:
\[ S = \sum_{r=1}^{30} \frac{{{30+r} \choose r} (2r-1)}}{{{30} \choose r} (30+r)} \]
We can rewrite the summation starting from \( r=0 \) to \( r=30 \):
\[ S = \sum_{r=0}^{30} \frac{{{30+r} \choose r} (2r-1)}}{{{30} \choose r} (30+r)} - \text{(the term for } r=0\text{)} \]
The term for \( r=0 \) is zero since \( (2*0 - 1) = -1 \).

2. **Simplifying \( S \)**:
We can factor out \( \frac{{{30+r} \choose r}}{{{30} \choose r}} \) and rewrite the summation:
\[ S = \sum_{r=0}^{30} \frac{{{30+r} \choose r}}{{{30} \choose r}} - \sum_{r=0}^{30} \frac{{{30+r} \choose r} (r-1)}{{{30} \choose r} (30+r)} \]
The first term simplifies to \( 2^{30} \) by the binomial theorem.

3. **Evaluate \( K \)**:
\[ K = \sum_{r=0}^{30} \left( {30 \choose r} \right)^2 \]
By the identity of binomial coefficients, this can be simplified using:
\[ K = {60 \choose 30} \]

4. **Evaluate \( G \)**:
\[ G = \sum_{r=0}^{60} (-1)^r \left( {60 \choose r} \right)^2 \]
This can be evaluated using the identity:
\[ G = {60 \choose 30} \]
This is because the sum of squares of binomial coefficients with alternating signs gives the central binomial coefficient.

5. **Combine \( K \) and \( G \)**:
Now we can find \( K + G \):
\[ K + G = {60 \choose 30} + {60 \choose 30} = 2 {60 \choose 30} \]

### Final Result:
The value of \( K + G \) is:
\[ K + G = 2 {60 \choose 30} \]