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What is the constant term in the expansion of $\left(\sqrt{x}+\dfrac5x\right)^{9}$?

Apr 23, 2021

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:
$${9 \choose 6}(\sqrt{x})^6(\frac{5}{x})^3={9 \choose 6}x^3(\frac{125}{x^3})={9 \choose 6}125=84\cdot125=\boxed{10500}$$