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The greatest coefficient in ((x^(3//2)y)...

The greatest coefficient in `((x^(3//2)y)/(2) + (2)/(xy^(3//2)))^12` is

A

`12(2^11)`

B

`12(2^10)`

C

`12(2^22)`

D

`33(2^9)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the greatest coefficient in the expression \(\left(\frac{x^{3/2}y}{2} + \frac{2}{xy^{3/2}}\right)^{12}\), we will use the Binomial Theorem. Here are the steps to solve the problem: ### Step 1: Identify the terms in the binomial expansion The expression can be rewritten as: \[ \left(a + b\right)^{n} \] where \(a = \frac{x^{3/2}y}{2}\) and \(b = \frac{2}{xy^{3/2}}\), and \(n = 12\). ### Step 2: Apply the Binomial Theorem According to the Binomial Theorem, the expansion of \((a + b)^n\) is given by: \[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \] ### Step 3: Find the general term The general term \(T_k\) in the expansion can be expressed as: \[ T_k = \binom{n}{k} a^{n-k} b^{k} \] Substituting the values of \(a\) and \(b\): \[ T_k = \binom{12}{k} \left(\frac{x^{3/2}y}{2}\right)^{12-k} \left(\frac{2}{xy^{3/2}}\right)^{k} \] ### Step 4: Simplify the general term Now simplify \(T_k\): \[ T_k = \binom{12}{k} \left(\frac{1}{2}\right)^{12-k} \left(\frac{2^k}{x^k y^{3k/2}}\right) \left(x^{(3/2)(12-k)} y^{(12-k)}\right) \] Combining the terms: \[ T_k = \binom{12}{k} \cdot \frac{2^k}{2^{12-k}} \cdot x^{(3/2)(12-k) - k} \cdot y^{(12-k) - \frac{3k}{2}} \] This simplifies to: \[ T_k = \binom{12}{k} \cdot \frac{2^k}{2^{12}} \cdot x^{18 - \frac{5k}{2}} \cdot y^{12 - \frac{5k}{2}} \] ### Step 5: Determine the exponent conditions To find the greatest coefficient, we need to find the value of \(k\) that maximizes the coefficient. The coefficient of \(T_k\) is: \[ \text{Coefficient} = \binom{12}{k} \cdot \frac{2^k}{2^{12}} = \binom{12}{k} \cdot \frac{1}{2^{12-k}} \] ### Step 6: Find the maximum coefficient The maximum coefficient occurs at \(k = \frac{n}{2} = \frac{12}{2} = 6\). Thus, we need to calculate: \[ \binom{12}{6} \] ### Step 7: Calculate \(\binom{12}{6}\) Using the formula for combinations: \[ \binom{12}{6} = \frac{12!}{6!6!} = \frac{479001600}{720 \cdot 720} = 924 \] ### Final Answer The greatest coefficient in the expansion is: \[ \boxed{924} \]
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