The number of integral terms in the expansion of ( 5 1 2 + 7 1 6 ) 642 is Options: A . &nbsp 106 B . &nbsp 108 C . &nbsp 103 D . &nbsp 109

Answer: Option B : B T r + 1 = 642 C r ( 5 1 2 ) 642 − r . ( 7 1 6 ) r Obviously, r should be a multiple of 6. Total number of terms = 642 6 = 107, but first term for r = 0 is also an integer. Hence total terms are 107 + 1 = 108.

The number of integral terms in the expansion of (512+716)642 is

Discussion Forum : Binomial Theorem

Question -

The number of integral terms in the expansion of $(5^{\frac{1}{2}} + 7^{\frac{1}{6}})^{642}$ is

Options:

A . 106

B . 108

C . 103

D . 109

Answer: Option B
:
B

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

Obviously, r should be a multiple of 6.

Total number of terms = $\frac{642}{6}$ = 107,

but first term for r = 0 is also an integer. Hence total terms are 107 + 1 = 108.

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Discussion Forum : Binomial Theorem

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