Question

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

• A. 32
• B. 33
• C. 34
• D. 35

Answer 1

The correct option is B 33

$(\sqrt{3}{+}^8\sqrt{5}{)}^{256}\therefore T_{r+1}{=}^{256}{C}_r(\sqrt{3}{)}^{256-r}{(}^8\sqrt{5}{)}^r{=}^{256}{C}_r\left(3{)}^{\frac{256-r}{2}}\right(5{)}^{\frac{r}{8}}$
Terms would be integeral if $\frac{(256-r)}{2}$ And $\frac{r}{8}$ are possitive integers.
As $0\le r\le 256$
$r=0,8,16,24\cdots 256$
For the above values of r, $\frac{(256-r)}{2}$ is also an integer. Hence, the total number fo values of r is 33.