The number of integral terms in the expansion | vedclass

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(B) The general term in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (3^{1/2})^{256-r} (5^{1/8})^r$. For the ter

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(B) The general term in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (3^{1/2})^{256-r} (5^{1/8})^r$. For the term to be an integer,the powers of $3$ and $5$ must be integers. This implies $\frac{256-r}{2}$ and $\frac{r}{8}$ must be integers. Since $r$ must be a multiple of $8$,let $r = 8k$,where $k \in \{0, 1, 2, \dots, 32\}$. Substituting $r = 8k$ into the expression for the power of $3$: $\frac{256-8k}{2} = 128-4k$,which is always an integer for any integer $k$. Thus,$r$ can take values $0, 8, 16, \dots, 256$. This is an arithmetic progression with first term $a = 0$,common difference $d = 8$,and last term $l = 256$. Using the formula $l = a + (n-1)d$,we get $256 = 0 + (n-1)8$. $32 = n-1$,which gives $n = 33$.

The number of integral terms in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$ is

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