The coefficient of (r + 1)th term in the expansion of `(1+x)^(n+1)` is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of `(1 + x)^(n)`.

Answer

To solve the problem, we need to find the coefficients of specific terms in the binomial expansions and establish the relationship stated in the question.

### Step-by-Step Solution:

1. **Identify the Coefficient of the (r + 1)th Term in (1 + x)^(n + 1)**:
The (r + 1)th term in the expansion of (1 + x)^(n + 1) can be found using the binomial theorem:
\[ T_{r+1} = \binom{n+1}{r} x^r \]
Therefore, the coefficient of the (r + 1)th term is:
\[ C_{r+1} = \binom{n+1}{r} \]

2. **Identify the Coefficients of the rth and (r + 1)th Terms in (1 + x)^(n)**:
Using the binomial theorem again, the rth and (r + 1)th terms in the expansion of (1 + x)^(n) are:
\[ T_r = \binom{n}{r} x^r \quad \text{and} \quad T_{r+1} = \binom{n}{r+1} x^{r+1} \]
Thus, the coefficients are:
\[ C_r = \binom{n}{r} \quad \text{and} \quad C_{r+1} = \binom{n}{r+1} \]

3. **Sum of the Coefficients of the rth and (r + 1)th Terms in (1 + x)^(n)**:
The sum of the coefficients of the rth and (r + 1)th terms is:
\[ S = C_r + C_{r+1} = \binom{n}{r} + \binom{n}{r+1} \]

4. **Apply Pascal's Identity**:
According to Pascal's identity, we have:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]
Therefore, we can rewrite the sum of the coefficients as:
\[ S = \binom{n+1}{r+1} \]

5. **Establish the Relationship**:
Now we can equate the coefficients we found:
\[ C_{r+1} = S \]
This gives us:
\[ \binom{n+1}{r} = \binom{n+1}{r+1} \]
This confirms the relationship stated in the question.

### Conclusion:
The coefficient of the (r + 1)th term in the expansion of (1 + x)^(n + 1) is indeed equal to the sum of the coefficients of the rth and (r + 1)th terms in the expansion of (1 + x)^(n).