The number of integral terms in the expansion | vedclass

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(B) The general term in the expansion of $(3^{1/2} + 5^{1/4})^{680}$ is given by $T_{r+1} = {}^{680}C_r (3^{1/2})^{680-r} (5^{1/4})^r$.This simplifies

Vedclass · Accepted Answer

$171$

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(B) The general term in the expansion of $(3^{1/2} + 5^{1/4})^{680}$ is given by $T_{r+1} = {}^{680}C_r (3^{1/2})^{680-r} (5^{1/4})^r$.This simplifies to $T_{r+1} = {}^{680}C_r \cdot 3^{(680-r)/2} \cdot 5^{r/4}$.For the term to be an integer,both exponents must be integers.$1$) $r/4$ must be an integer,so $r$ must be a multiple of $4$. Thus,$r \in \{0, 4, 8, \dots, 680\}$.$2$) $(680-r)/2$ must be an integer,which means $680-r$ must be even. Since $680$ is even,$r$ must be even.Since all multiples of $4$ are even,the condition $r \in \{0, 4, 8, \dots, 680\}$ satisfies both requirements.The number of such terms is the number of values in the sequence $0, 4, 8, \dots, 680$.Using the formula for the number of terms in an arithmetic progression,$n = \frac{last - first}{difference} + 1 = \frac{680 - 0}{4} + 1 = 170 + 1 = 171$.

The number of integral terms in the expansion of $(3^{1/2} + 5^{1/4})^{680}$ is equal to

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