写出并化简 $\left(2 - x\right)^7$ 按 $x$ 的升幂排列的展开式中的前 3 项。 求 $\left(1 - 8x + 24x^2\right)\left(2 - x\right)^7$ 展开式中 $x^2$ 的系数。 利用通项，求 $\left(2x - \dfrac{1}{x^2}\right)^{17}$ 的二项式展开式中含 $\frac{1}{x^7}$ 的项。 解释为什么在 $\left(2x - \dfrac{1}{x^2}\right)^{17}$ 的展开式中没有含 $\dfrac{1}{x^2}$ 的项。

1. Apply the binomial theorem to expand the first three terms

$\left(2 - x\right)^7 = 2^7 + \binom{7}{1}2^6(-x) + \binom{7}{2}2^5(-x)^2 + \dots$

2. Simplify the terms

$= 128 + 7(64)(-x) + 21(32)(x^2) + \dots$
$= 128 - 448x + 672x^2$

1. Recall or calculate the first 3 terms of the expansion of $(2 - x)^7$ from part (a)(i)

From part (a)(i),
$(2 - x)^7 = 2^7 + \binom{7}{1}2^6(-x) + \binom{7}{2}2^5(-x)^2 + \dots$
$= 128 - 448x + 672x^2 + \dots$

2. Identify the combinations of terms that produce $x^2$ in the product

We want the coefficient of $x^2$ in:
$(1 - 8x + 24x^2)(128 - 448x + 672x^2 + \dots)$
The terms in $x^2$ are obtained by multiplying:

• The constant term in the first bracket with the $x^2$ term in the second bracket
• The $x$ term in the first bracket with the $x$ term in the second bracket
• The $x^2$ term in the first bracket with the constant term in the second bracket

3. Calculate the coefficient of $x^2$

Coefficient of $x^2 = (1)(672) + (-8)(-448) + (24)(128)$
$= 672 + 3584 + 3072$
$= 7328$

1. Write down the general term for the expansion

The general term is given by:
$T_{r+1} = \binom{17}{r} (2x)^{17-r} \left(-\frac{1}{x^2}\right)^r$
$T_{r+1} = \binom{17}{r} 2^{17-r} (-1)^r x^{17-r-2r}$
$T_{r+1} = \binom{17}{r} 2^{17-r} (-1)^r x^{17-3r}$

2. Find the value of $r$ for the term in $\frac{1}{x^7}$

For the term in $\frac{1}{x^7}$, the power of $x$ must be $-7$.
$17 - 3r = -7$
$3r = 24$
$r = 8$

3. Calculate the term by substituting $r = 8$

Substitute $r = 8$ into the general term:
$T_9 = \binom{17}{8} 2^{17-8} (-1)^8 x^{-7}$
$T_9 = 24310 \times 2^9 \times 1 \times x^{-7}$
$T_9 = 24310 \times 512 \times x^{-7}$
$T_9 = \frac{12446720}{x^7}$

1. Equate the power of $x$ in the general term to $-2$

From the general term $T_{r+1} = \binom{17}{r} 2^{17-r} (-1)^r x^{17-3r}$, we require the power of $x$ to be $-2$.
$17 - 3r = -2$

2. Solve for $r$ and conclude based on the integer requirement

$3r = 19$
$r = \frac{19}{3} = 6\frac{1}{3}$
Since $r$ must be an integer ($0 \leqslant r \leqslant 17$), and $6\frac{1}{3}$ is not an integer, there is no term in $\frac{1}{x^2}$.