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The number of terms with integral coefficients in the expansion of $(17^\frac{1}{3}+19^\frac{1}{2}x)^{600}$ is

  1. $99$
  2. $100$
  3. $101$
  4. $102$
in Combinatory by Boss (16.8k points)
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1 Answer

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$\underline{\mathbf{Answer: C}}$

$\underline{\mathbf{Explanation:}}$

$\begin {align} \text{Number of terms in the expansion }({17}^{\frac{1}{3}} + 19^{\frac{1}{2}}\mathrm x)&=({17}^{\frac{1}{3}} + 19^{\frac{1}{2}}\mathrm x)\\&=600_{C_r}17^{\frac{600-r}{3}}.19^{\frac{r}{2}}.x^r\\&=600_{C_r}(17^{\frac{1}{3}})^{\mathrm A}(19^{\frac{1}{2}})^{\mathrm B}.x^{\mathrm B}\end {align}$

$\text{Now}, \mathrm A + \mathrm B = 600$

 

$\begin{align} \text{Now A and B should be a multiple of 3 and 2 respectively.}\\ \mathrm A = 3\lambda, \; \lambda \ge0, \mathrm B = 2\mu, \;\mu \ge0\end{align}$

 

$\text{Now}, \underbrace {3\lambda}_\text {=Even} + \underbrace{2\mu}_\text {=Even} = 600$

$\text {Now if, }\lambda = 0, 2, 4, 6,\cdots\cdots,200\;\text {are in AP}$

$\text{Then, }\; \mathrm {T_n} = a + (n-1)d = 200 \Rightarrow (0+(n-1)2) = 200 \Rightarrow n = 101$

$\therefore $ is the correct option.

by Boss (13.2k points)
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