If the coefficient of $x^r$ in $(1 + x)^n$ equals the coefficient of $x^{r+2}$ in the same expansion, then

1. Coefficients in $(1+x)^n$: of $x^r$ is $\binom{n}{r}$; of $x^{r+2}$ is $\binom{n}{r+2}$. 2. Equality: $\binom{n}{r} = \binom{n}{r+2}$. 3. Use the identity $\binom{n}{k} = \binom{n}{n-k}$: so $\binom{n}{r} = \binom{n}{n-r}$. 4. For $\binom{n}{r} = \binom{n}{r+2}$, we need $r+2 = n-r$ (the only way two different lower-index choices give the same value, assuming neither is the symmetric partner of itself). 5. So $n = 2r + 2$. 6. Example: $n = 6$, $r = 2$, $r+2 = 4$. $\binom{6}{2} = 15 = \binom{6}{4} = 15$ ✓. 7. Other options misapply the symmetry argument. _Source: NCERT Class 11 Mathematics, Ch 7, §7.2.1 + §7.2.2 (Symmetry of coefficients), p. 5._