The number of integral terms in the expansion of

$(5^{\frac{1}{2}} + 7^{\frac{1}{6}})^{642}$ is

106

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108

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103

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109

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Solution

The correct option is B
108

$T_{r + 1}$ = $^{642} C_{r} (5^{\frac{1}{2}})^{642 - r} . (7^{\frac{1}{6}})^{r}$

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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