What is the constant term in the expansion of $\left(\sqrt{x}+\dfrac5x\right)^{9}$?
To get the constant term, the $\sqrt{x}$'s and $5/x$'s must cancel out. This occurs for the term with six $\sqrt{x}$'s and three $5/x$'s, so the coefficient is $$\binom93\cdot5^3=84\cdot125=\boxed{10500}.$$
10500
Level 4
Counting & Probability
Problem: What is the constant term in the expansion of $\left(\sqrt{x}+\dfrac5x\right)^{9}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the constant term in the expansion of \((\sqrt{x} + \frac{5}{x})^9\), we apply the binomial theorem, which states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = \sqrt{x}\), \(b = \frac{5}{x}\), and \(n = 9\).

The general term in the expansion can be expressed as:
\[
T_k = \binom{9}{k} (\sqrt{x})^{9-k} \left(\frac{5}{x}\right)^k
\]
This simplifies to:
\[
T_k = \binom{9}{k} \cdot 5^k \cdot (\sqrt{x})^{9-k} \cdot x^{-k}
\]
Next, we rewrite \((\sqrt{x})^{9-k}\) as \(x^{(9-k)/2}\). Thus, the term becomes:
\[
T_k = \binom{9}{k} \cdot 5^k \cdot x^{(9-k)/2 - k}
\]
We need to find the value of \(k\) such that the exponent of \(x\) is zero, which gives us the constant term. Setting the exponent equal to zero:
\[
\frac{9-k}{2} - k = 0
\]
Multiplying through by 2 to eliminate the fraction:
\[
9 - k - 2k = 0 \implies 9 - 3k = 0 \implies 3k = 9 \implies k = 3
\]
Now, substituting \(k = 3\) back into the expression for \(T_k\):
\[
T_3 = \binom{9}{3} \cdot 5^3 \cdot x^{0}
\]
Calculating \(\binom{9}{3}\):
\[
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
\]
Calculating \(5^3\):
\[
5^3 = 125
\]
Thus, the constant term is:
\[
T_3 = 84 \cdot 125 = 10500
\]
The final answer is \(10500\).<|END_OF_LATENT|> The final answer is $10500$.