What is the constant term in the expansion of $\left(\sqrt{x}+\dfrac5x\right)^{9}$?
+505 First of all, you should be familiar with the binomial theorem, which is: \(\displaystyle (a+b)^n = \sum_{k=0}^n {n\choose k}a^kb^{n-k}\). In this case, \(a=\sqrt{x}, b=\frac{5}{x}\), and \(n=9\). The value of k that is going to produce the constant term is 6, because the x term in a becomes x^3 and the x term in b becomes 1/x^3, and when multiplied it cancels out. That constant term will look like this:
\({9 \choose 6}(\sqrt{x})^6(\frac{5}{x})^3={9 \choose 6}x^3(\frac{125}{x^3})={9 \choose 6}125=84\cdot125=\boxed{10500}\)
