The number of terms with integral co-efficients in the expansion of $${\left( {{7^{\frac{1}{3}}} + {5^{\frac{1}{2}}} \cdot x} \right)^{600}}$$ is
A.
100
B.
50
C.
101
D.
None of these
Answer :
101
Solution :
$${t_{r + 1}} = {\,^{600}}{C_r} \cdot {7^{\frac{{600 - r}}{3}}} \cdot {5^{\frac{r}{2}}}{x^r}.$$
Here, $$0 \leqslant r \leqslant 600$$ and $$\frac{r}{2},200 - \frac{r}{3}$$ are integers.
∴ $$r$$ should be a multiple of 6
$$\therefore \,\,r = 0,6,12,.....,600.$$
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