Question
The number of integral terms in the expansion of $${\left( {\sqrt 3 + \root 8 \of 5 } \right)^{256}}$$ is
A.
35
B.
32
C.
33
D.
34
Answer :
33
Solution :
$$\eqalign{
& {T_{r + 1}} = {\,^{256}}{C_r}{\left( {\sqrt 3 } \right)^{256 - r}}{\left( {^8\sqrt 5 } \right)^r} \cr
& \,\,\,\,\,\,\, = {\,^{256}}{C_r}{\left( 3 \right)^{\frac{{256 - r}}{2}}}{\left( 5 \right)^{\frac{r}{8}}} \cr} $$
Terms will be integral if $$\frac{{256 - r}}{2}\& \frac{r}{8}$$ both are +ve
integer, which is so if $$r$$ is an integral multiple of 8. As $$0 \leqslant r \leqslant 256$$
∴ $$r$$ = 0, 8, 16, 24, . . . . . , 256, total 33 values.