The number of rational terms in the binomial | vedclass

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(B) The general term $T_{r+1}$ in the expansion of $(4^{1/4} + 5^{1/6})^{120}$ is given by:$T_{r+1} = {}^{120}C_r (4^{1/4})^{120-r} (5^{1/6})^r$$T_{r+

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(B) The general term $T_{r+1}$ in the expansion of $(4^{1/4} + 5^{1/6})^{120}$ is given by:$T_{r+1} = {}^{120}C_r (4^{1/4})^{120-r} (5^{1/6})^r$$T_{r+1} = {}^{120}C_r (2^{2/4})^{120-r} (5^{r/6})$$T_{r+1} = {}^{120}C_r (2^{1/2})^{120-r} (5^{r/6})$$T_{r+1} = {}^{120}C_r (2^{60 - r/2}) (5^{r/6})$For the term to be rational,the exponents of $2$ and $5$ must be integers.Thus,$r/2$ must be an integer (so $r$ is a multiple of $2$) and $r/6$ must be an integer (so $r$ is a multiple of $6$).Therefore,$r$ must be a multiple of $	ext{lcm}(2, 6) = 6$.Given $0 \leq r \leq 120$,the possible values for $r$ are $0, 6, 12, \dots, 120$.This is an arithmetic progression where $a = 0$,$d = 6$,and $l = 120$.The number of terms $n$ is given by $120 = 0 + (n-1)6$,which gives $n-1 = 20$,so $n = 21$.Thus,there are $21$ rational terms.

The number of rational terms in the binomial expansion of $(4^{1/4} + 5^{1/6})^{120}$ is $....$

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