A) \[\frac{160}{9}\]
B) \[\frac{80}{9}\]
C) \[\frac{160}{27}\]
D) \[\frac{80}{3}\]
Correct Answer: C
Solution :
| Given expansion \[{{\left( 2x+\frac{1}{3x} \right)}^{6}}\] |
| Let \[(r+1)th\] term be the independent of x. |
| \[(r+1)term{{=}^{6}}{{C}_{r}}{{(2x)}^{6-r}}{{\left( \frac{1}{3x} \right)}^{r}}\]\[{{=}^{6}}{{C}_{r}}{{2}^{6-r}}{{x}^{6-r}}\frac{1}{{{3}^{r}}{{x}^{r}}}\] |
| \[{{=}^{6}}{{C}_{r}}\,{{2}^{6-r}}.\,{{3}^{-r}}.{{x}^{6-2r}}\] ?..(1) |
| So, \[{{x}^{0}}={{x}^{6}}-2r\,\,\Rightarrow \,\,6-2r=0\] |
| \[2r=6\Rightarrow r=3\] |
| Taking eq. (1), we get \[{{(3+1)}^{th}}term\,\,{{=}^{6}}{{C}_{3}}{{(2)}^{6-3}}{{3}^{-3}}.{{x}^{0}}=\frac{6\times 5\times 4}{3\times 2}\times \frac{{{2}^{3}}}{{{3}^{3}}}\] \[{{(4)}^{th}}term=\frac{160}{27}\] |
You need to login to perform this action.
You will be redirected in
3 sec
