Let the coefficient of three consecutive terms \({T}_{r},{T}_{r+1},&{T}_{r+2}\)in the binomial expansion of \({(a+b)}^{1…
Let the coefficient of three consecutive terms \({T}_{r},{T}_{r+1},&{T}_{r+2}\)in the binomial expansion of \({(a+b)}^{12}\) be in a A.P. and let p be the no. of all possible values of r, let q be the sum of all rational term in the binomial expansion \({(\sqrt[4]{3}+\sqrt[3]{4})}^{12}\). then p+q is equal to:
283
\({(\sqrt[4]{3}+\sqrt[3]{4})}^{12}\)
| Exponent of \(\sqrt[4]{3}\) | Exponent of \(\sqrt[3]{4}\) | Term |
| 12 | 0 | 27 |
| 0 | 12 | 256 |
q = 27+256 = 283
Now, \({}^{12}\mathrm{C}_{\mathrm{r}-1}+{}^{12}\mathrm{C}_{\mathrm{r}+1}=2\cdot {}^{12}\mathrm{C}_{\mathrm{r}}\)
\(\frac{12!}{(r-1)!(13-r)!}+\frac{12!}{(r+1)!(11-r)!}=2\cdot \frac{12!}{r!(12-r)!}\)
\(\frac{1}{(13-r)(12-r)}+\frac{1}{(r+1)(r)}=\frac{2}{r(12-r)}\)
\(2{r}^{2}-24r+156=26r+26-2{r}^{2}-2r\)
\(2{\mathrm{r}}^{2}-24\mathrm{r}+65=0\)
No integral value of r.
\(p=0\\ p+q=283\)
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