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 :
$${T_{r + 1}} = {\,^{256}}{C_r}{\left( {\sqrt 3 } \right)^{256 - r}}{\left( {\root 8 \of 5 } \right)^r} = {\,^{256}}{C_r}{\left( 3 \right)^{\frac{{256 - r}}{2}}}{\left( 5 \right)^{\frac{r}{8}}}$$
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 $$0 \leqslant r \leqslant 256$$
$$\therefore r = 0,8,16,24,.....256,{\text{ total }}33{\text{ values}}{\text{.}}$$
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is