A) 40
B) 41
C) 42
D) 0
Correct Answer: B
Solution :
| [b] Here \[{{T}_{r+1}}={{\,}^{10}}{{C}_{r}}{{(\sqrt{2})}^{10-r}}{{({{3}^{1/5}})}^{r}},\] |
| where r = 0, 1, 2, ...., 10. |
| We observe that in general term \[{{T}_{r+1}}\] powers of 2 and 3 are \[\frac{1}{2}(10-r)\] and \[\frac{1}{5}r\] respectively and 0 |
| \[\le r\le 10\]. |
| So, both these powers will be integers together only when \[r=0\] or 10 |
| \[\therefore \] Sum of required terms \[={{T}_{1}}+{{T}_{11}}\] |
| \[={{\,}^{10}}{{C}_{0}}{{(\sqrt{2})}^{10}}+{{\,}^{10}}{{C}_{10}}\,{{({{3}^{1/5}})}^{10}}=32+9=41\] |
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