Let the coefficients of three consecutive terms $T_r, T_{r+1}$ and $T_{r+2}$ in the binomial expansion of…
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 $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $\mathrm{p}+\mathrm{q}$ is equal to :
$283$
$287$
$295$
$299$
Solution
Coefficient of
$\begin{aligned}
& T_r, T_{r+1}, T_{r+2} \rightarrow G P \\
& \Rightarrow\left({ }^{12} C_r\right)^2={ }^{12} C_{r-1} \cdot{ }^{12} C_{r+1}
\end{aligned}$
$\Rightarrow\left({ }^{12} C_r\right)^2={ }^{12} C_{r-1} \cdot{ }^{12} C_{r+1}$
but no three consecutive binomial coefficient are in
GP
$\Rightarrow P=0$
Now for $\left(3^{1 / 4}+4^{1 / 3}\right)^{12}, T_{r+1}={ }^{12} C_r(4)^{K / 3}(3)^{\frac{12-K}{4}}$
for rational terms $\mathrm{K}=0,12$
sum of rational terms
$\begin{aligned}
& ={ }^{12} \mathrm{C} 04^0 \cdot 3^3+{ }^{12} \mathrm{C}_{12} \cdot 4^4 \cdot 3^0 \\
& =27+256=283=q \\
& \therefore p+q=283
\end{aligned}$