A) 106
B) 108
C) 103
D) 109
Correct Answer: B
Solution :
\[{{T}_{r+1}}{{=}^{642}}{{C}_{r}}{{({{5}^{1/2}})}^{642-r}}.\,{{({{7}^{1/6}})}^{r}}\] Obviously, r should be a multiple of 6. Total numbers = \[\frac{642}{6}=107\]; But first term for \[r=0\] is also integral. Hence total terms are\[107+1=108\].
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