The number of integral terms in the expansion | vedclass

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(B) The general term of the expansion is given by $T_{r+1} = {}^{642}C_r (5^{1/2})^{642-r} (7^{1/6})^r$.For the term to be integral,the powers of $5$

Vedclass · Accepted Answer

$108$

Vedclass · Answer

(B) The general term of the expansion is given by $T_{r+1} = {}^{642}C_r (5^{1/2})^{642-r} (7^{1/6})^r$.For the term to be integral,the powers of $5$ and $7$ must be integers.Since $5^{(642-r)/2} = 5^{321 - r/2}$,$r$ must be an even integer.Since $7^{r/6}$ must be an integer,$r$ must be a multiple of $6$.Combining these,$r$ must be a multiple of $6$ in the range $0 \le r \le 642$.The possible values for $r$ are $0, 6, 12, \dots, 642$.This is an arithmetic progression where $a = 0$,$d = 6$,and $l = 642$.The number of terms $n$ is given by $642 = 0 + (n-1)6$,which gives $n-1 = 107$,so $n = 108$.

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

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