Write value of
` "^(2n-1)C_(5)+"^(2n-1)C_(6)+"^(2n)C_(7)` use `["^(n)c_(r)+^(n)C_(r-1)="^(n+1)C_(r)]`

Answer

To solve the expression \( \binom{2n-1}{5} + \binom{2n-1}{6} + \binom{2n}{7} \), we will use the identity:

\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \]

### Step-by-step Solution:

1. **Identify the terms**: We have the expression \( \binom{2n-1}{5} + \binom{2n-1}{6} + \binom{2n}{7} \).

2. **Combine the first two terms**: Using the identity, we can combine \( \binom{2n-1}{5} \) and \( \binom{2n-1}{6} \):
\[ \binom{2n-1}{5} + \binom{2n-1}{6} = \binom{2n}{6} \]
This is because \( n = 2n-1 \) and \( r = 6 \).

3. **Rewrite the expression**: Now, substitute back into the original expression:
\[ \binom{2n}{6} + \binom{2n}{7} \]

4. **Apply the identity again**: Now we can apply the identity again to combine these two terms:
\[ \binom{2n}{6} + \binom{2n}{7} = \binom{2n+1}{7} \]

5. **Final expression**: Therefore, the value of the original expression is:
\[ \binom{2n+1}{7} \]

### Final Answer:
\[ \binom{2n+1}{7} \]