Prove that `""^(n)C_(r )+2""^(n)C_(r-1)+ ""^(n)C_(r-2)= ""^(n+2)C_(r )`.

Answer

To prove the identity \( \binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r} \), we will manipulate the left-hand side (LHS) and show that it equals the right-hand side (RHS).

### Step-by-Step Solution:

1. **Start with the Left-Hand Side (LHS)**:
\[ LHS = \binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-2} \]

2. **Use the Identity for Combinations**:
We know the identity:
\[ \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} \]
We can apply this identity to the first two terms of the LHS:
\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \]
So now we can rewrite the LHS as:
\[ LHS = \binom{n+1}{r} + \binom{n}{r-2} \]

3. **Apply the Identity Again**:
Now we will apply the identity again to combine \( \binom{n+1}{r} \) and \( \binom{n}{r-2} \):
\[ \binom{n+1}{r} + \binom{n}{r-2} = \binom{n+2}{r} \]
This is because we can think of \( \binom{n}{r-2} \) as \( \binom{n+1}{r-1} + \binom{n+1}{r-2} \).

4. **Final Result**:
Therefore, we have:
\[ LHS = \binom{n+2}{r} \]
which is equal to the right-hand side (RHS):
\[ RHS = \binom{n+2}{r} \]

5. **Conclusion**:
Since \( LHS = RHS \), we have proved the identity:
\[ \binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r} \]