Let the coefficients of three consecutive terms T_r,T_r+1,T_r+2 in the binomial expansion …

Question

Let the coefficients of three consecutive terms $T_r,T_{r+1},T_{r+2}$ 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 $\left(\sqrt[4]3+\sqrt[3]4\right)^{12}$. Then $p+q$ is equal to:

Accepted Answer

$$\begin{aligned}
& (\mathrm{a}+\mathrm{b})^{\frac{1}{2}} \
& \mathrm{~T}_{\mathrm{r}}, \mathrm{~T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP} \
& \text { So, } \frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{\mathrm{T}_{\mathrm{r}+2}}{\mathrm{~T}_{\mathrm{r}+1}} \
& \frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}} \
& \frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1} \
& (13-\mathrm{r})(\mathrm{r}+1)=(12-\mathrm{r})(\mathrm{r}) \
& -\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^2 \
& 13=0
\end{aligned}$$

No value of r possible

So $\mathrm{P}=0$

$$\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C_r\left(3^{\frac{1}{4}}\right)^{12-\mathrm{r}}\left(4^{\frac{1}{3}}\right)^{\mathrm{r}}$$

Exponent of $\left(3^{\frac{1}{4}}\right)$ exponent of $\left(4^{\frac{1}{3}}\right)$ term

$$\[\begin{array}{ccc} 12 & 0 & 27 \ 0 & 12 & 256 \ \mathrm{q}=27+256=283 & & \ \mathrm{p}+\mathrm{q}=0+283=283 & & \end{array}$$\]

Solutions · 1