Question

determine the twelfth term of the following expression without expanding the binomial.

\\(3x - y)^{16}\\

Explanation:

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<reasoning>

Identify binomial parameters and term index

\[

$$\begin{aligned} &(a + b)^n = (3x - y)^{16} \\ &a = 3x, \quad b = -y, \quad n = 16 \\ &T_{k+1} = \binom{n}{k} a^{n-k} b^k \\ &k + 1 = 12 \implies k = 11 \end{aligned}$$

\]

Calculate binomial coefficient and substitute values

\[

$$\begin{aligned} \binom{16}{11} &= \frac{16!}{11! \cdot 5!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 4368 \\ T_{12} &= 4368 \cdot (3x)^{16-11} \cdot (-y)^{11} \\ T_{12} &= 4368 \cdot (3x)^5 \cdot (-y)^{11} \end{aligned}$$

\]

Simplify the expression

\[

$$\begin{aligned} 3^5 &= 243 \\ T_{12} &= 4368 \cdot 243 \cdot x^5 \cdot (-1) \cdot y^{11} \\ T_{12} &= -1,061,424x^5y^{11} \end{aligned}$$

\]
</reasoning>

<answer>
\(-1,061,424x^5y^{11}\)
</answer>

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"Mathematics",
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</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

Identify binomial parameters and term index

\[

$$\begin{aligned} &(a + b)^n = (3x - y)^{16} \\ &a = 3x, \quad b = -y, \quad n = 16 \\ &T_{k+1} = \binom{n}{k} a^{n-k} b^k \\ &k + 1 = 12 \implies k = 11 \end{aligned}$$

\]

Calculate binomial coefficient and substitute values

\[

$$\begin{aligned} \binom{16}{11} &= \frac{16!}{11! \cdot 5!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 4368 \\ T_{12} &= 4368 \cdot (3x)^{16-11} \cdot (-y)^{11} \\ T_{12} &= 4368 \cdot (3x)^5 \cdot (-y)^{11} \end{aligned}$$

\]

Simplify the expression

\[

$$\begin{aligned} 3^5 &= 243 \\ T_{12} &= 4368 \cdot 243 \cdot x^5 \cdot (-1) \cdot y^{11} \\ T_{12} &= -1,061,424x^5y^{11} \end{aligned}$$

\]
</reasoning>

<answer>
\(-1,061,424x^5y^{11}\)
</answer>

<post_analysis>
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"question_type": "Free Response",
"knowledge_point": [
"Mathematics",
"Algebra",
"Binomial Theorem"
]
}
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