Answer

Solution

1. 1

   General term: $\binom{19}{k}x^{19-k}x^{-2k}=\binom{19}{k}x^{19-3k}$. Constant: $19-3k=0\Rightarrow k$ not integer. So there is no constant term --- but $19-3k=0$ has no integer solution, hence the constant term equals $\boxed{0}$. Let us re-examine the options:

2. 2
   • $171$ \quad (B) $19$ \quad (C) $1$ \quad (D) none of the above
3. 3

   $19-3k=0$ gives $k=19/3\notin\mathbb Z$. So no constant term exists, i.e.\ the constant is $0$.

4. 4

   Wait --- with $k=6$: $19-18=1$, with $k=7$: $19-21=-2$. So indeed no $k\in\{0,\dots,19\}$ produces $x^0$.

5. 5

   However, the intended reading might be $\bigl(x+\tfrac{1}{x^{2}}\bigr)$, giving exponent $19-k+(-2)k=19-3k$, same conclusion. Answer: $0$, not in the listed numerical options.

Answer structure / marking notes

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