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 $(\sqrt[3]{4}+\sqrt[4]{3})^{12}$. Then $p+q$ is equal to:
Step-by-Step Solution
Key Concept: Three consecutive binomial coefficients in $(a+b)^{n}$ in G.P.\ force the cross-equation $(n-r+1)(r+1)=r(n-r)$, which simplifies to $n+1=0$ — impossible for $n\in\mathbb{N}$, so $p=0$. For rational terms in $(2^{2/3}+3^{1/4})^{12}$, need $3\mid(12-r)$ and $4\mid r$.
\textbf{G.P.\ part.} $\dfrac{\binom{12}{r}}{\binom{12}{r-1}}=\dfrac{\binom{12}{r+1}}{\binom{12}{r}}$ gives $\dfrac{13-r}{r}=\dfrac{12-r}{r+1}.$ Cross-multiplying: $13r+13-r^{2}-r=12r-r^{2}$, so $13=0$ — contradiction. Hence $p=0.$
\textbf{Rational terms.} $(4^{1/3}+3^{1/4})^{12}=(2^{2/3}+3^{1/4})^{12}.$ General term $\binom{12}{r}\,2^{2(12-r)/3}\,3^{r/4}.$ Rational iff $3\mid(12-r)$ AND $4\mid r$, i.e.\ $r\equiv 0\pmod{12}.$
With $0\le r\le 12$: $r=0,12.$
$r=0$: $\binom{12}{0}\,2^{8}=256.$
$r=12$: $\binom{12}{12}\,3^{3}=27.$
$q=256+27=283.$ Hence $p+q=0+283=283.$
Correct Answer: 1