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

  1. $99$
  2. $100$
  3. $101$
  4. $102$
in Combinatory by Boss (17.5k points)
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1 Answer

+1 vote
Best answer

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

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

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

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

 

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

 

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

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

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

$\therefore \mathbf C$ is the correct option.

by Boss (19.1k points)
edited by
0
Typo in line 1?
0
I don't think sir.

Its just I wrote one step directly.
+1
"600" missing?
0
Oh yes, Thank You Sir.
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