Q16 / 50 · Manipal (MET) 2024 · Mathematics

If the number of terms in the expansion of $(x \sqrt{180}+\sqrt[3]{432})^{200}$ having integral coefficients is $n$, then the vaule of $\left[\frac{n}{6}\right]$ is (where [.] denotes greater integer function)

  • A
    4
  • Correct answer
    B
    5
  • C
    6
  • D
    7

Solution

$\begin{aligned} & \text T_{r+1}={ }^{200} C_r(x \cdot \sqrt{180})^{200-r} \cdot(\sqrt[3]{432})^r \\ & ={ }^{200} C_r 2^{200-r} \cdot 3^{200-r} \cdot 5^{\frac{200-r}{2}} \cdot x^{200-r} 2^r 3^r 2^{\frac{r}{3}} \\ & \therefore \text { Coefficient }={ }^{200} C_r 2^{200} \cdot 3^{200} \cdot 5^{100} \cdot \frac{2^{\frac{r}{3}}}{5^{\frac{r}{2}}} \end{aligned}$ $r$ must be multiple of 6 and takes values between 0 to 200. $\therefore 34$ terms $=n$ $\left[\frac{n}{6}\right]=\left[\frac{34}{6}\right]=5$

Hence, the correct answer is option: 'B'.

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