A) six
B) twelve
C) seven
D) five
Correct Answer: A
Solution :
Given expansion is \[{{\left( {{y}^{{}^{1}/{}_{5}}}+{{x}^{{}^{1}/{}_{10}}} \right)}^{55}}\] The general term is \[{{T}_{r+1}}{{=}^{55}}{{C}_{r}}{{\left( {{y}^{{}^{1}/{}_{5}}} \right)}^{55-r}}{{\left( {{x}^{\frac{1}{10}}} \right)}^{r}}\] \[{{T}_{r+1}}\]would free from radical sign if powers of y and x are integers. i.e. \[\frac{55-r}{5}\] and \[\frac{r}{10}\] are integer. \[\Rightarrow \]r is multiple of 10. Hence, r = 0,10,20,30,40,50 It is an A.P. Thus, \[50=0+(k-1)10\] \[50=10k-10\Rightarrow k=6\] Thus, the six terms of the given expansion in which x and y are free from radical signs.
You need to login to perform this action.
You will be redirected in
3 sec
