Binomial Theorem
Rational Terms + GP Coefficients
nta_pyq_2025_apr
Grade 11
Question:
Let the coefficients of three consecutive terms $T_r, T_{r+1}$ and $T_{r+2}$ in the binomial expansion of $(a + b)^{12}$ be in a G.P. and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[3]{3} + \sqrt[4]{4})^{12}$. Then $p + q$ is equal to:
Step-by-Step Solution
Key Concept: No three consecutive binomial coefficients ${}^nC_{r-1}, {}^nC_r, {}^nC_{r+1}$ can be in GP (so $p = 0$). For rational terms in $(3^{1/3} + 4^{1/4})^{12}$, the general term is ${}^{12}C_K \cdot 4^{K/3} \cdot 3^{(12-K)/4}$; rationality requires $K = 0$ or $K = 12$.
$p = 0$ since no three consecutive binomial coefficients are in GP. For $(3^{1/4} + 4^{1/3})^{12}$, rational terms occur at $K = 0$ and $K = 12$: $q = {}^{12}C_0 \cdot 3^3 + {}^{12}C_{12} \cdot 4^4 \cdot 3^0 = 27 + 256 = 283$. So $p + q = 283$.
Correct Answer: 283