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 \((\sqrt[4]{3}+\sqrt[3]{4})^{12}.\) Then \(\mathrm{p}+\mathrm{q}\) is equal to :

(1) 283

(2) 295

(3) 287

(4) 299

Answer 1

Correct option is (1) 283

\((a+b)^{\frac{1}{2}}\)

\(\mathrm{T}_{\mathrm{r}}, \mathrm{T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP}\)

So, \(\frac{T_{r+1}}{T_{r}}=\frac{T_{r+2}}{T_{r+1}}\)

\(\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}}\)

\(\frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1}\)   

(13 - r) (r + 1) = (12 - r) (r)

\(-\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^{2}\)    

13=0

No value of r possible

So \(\mathrm{P}=0\)

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

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

12	0	27
0	12	256

q = 27 + 256 = 283

p + q = 0 + 283 = 283