Binomial Theorem
Binomial Theorem
star_batch_jee_advanced_2025
Grade 11
Question:
If $n \in \mathbb{N}$ and $(1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r$, then $\sum_{r=0}^{n} (-1)^r a_r \, ^nC_r$ is equal to:
0 if n = 57
0 if n = 77
^{24}C_8 if n = 24
^{39}C_{13} if n = 39
Step-by-Step Solution
Key Concept: The sum $\sum_{r=0}^{n} (-1)^r a_r \binom{n}{r}$ evaluates to zero when $n$ is odd due to the trinomial expansion $(1+x+x^2)^n$ and roots of unity filter properties.
We have $(1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r$. To find $\sum_{r=0}^{n} (-1)^r a_r \binom{n}{r}$, we use the substitution $x = -1$ in a related expression. Note that $\sum_{r=0}^{n} (-1)^r a_r \binom{n}{r}$ can be evaluated by considering $(1+x+x^2)^n$ and extracting coefficients through the binomial transform. Setting $x = -1$: $(1-1+1)^n = 1^n = 1$ when we work with the appropriate generating function. However, the sum $\sum_{r=0}^{n} (-1)^r a_r \binom{n}{r}$ equals $0$ when $n \equiv 0 \pmod{3}$ or when $n$ is odd, which occurs for $n = 57$ (odd) and $n = 77$ (odd). Checking divisibility: $57 = 3 \times 19$ and $77 = 7 \times 11$, so $n=77$ satisfies the condition.
Correct Answer: 2