在 (x+kx)5\left(x + \frac{k}{x}\right)^5 的二项式展开式中(其中 kk 是正整数),1x3\frac{1}{x^3} 和 1x\frac{1}{x} 的系数相同。求 kk 的值。 在不展开所有项的情况下,解释为什么 (1−3x2)(x+kx)5\left(1 - 3x^2\right)\left(x + \frac{k}{x}\right)^5 的展开式中没有常数项。 利用你的图表估算 pp 和 kk 的值, 35 分钟后液体的温度, 液体温度下降其初始值的 75% 所需的时间(分钟数)。 前 4 秒内运动的总路程。

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完整解法

6(a)

1. Write down the general term of the binomial expansion

The general term Tr+1T_{r+1} in the expansion of (x+kx)5\left(x + \frac{k}{x}\right)^5 is:
Tr+1=(5r)x5−r(kx)rT_{r+1} = \binom{5}{r} x^{5-r} \left(\frac{k}{x}\right)^r

2. Simplify the general term to isolate the power of x

Tr+1=(5r)krx5−rx−r=(5r)krx5−2rT_{r+1} = \binom{5}{r} k^r x^{5-r} x^{-r} = \binom{5}{r} k^r x^{5-2r}

3. Find the value of r for the term in 1/x^3

For the term in 1x3\frac{1}{x^3} (or x−3x^{-3}):
5−2r=−35 - 2r = -3
2r=8  ⟹  r=42r = 8 \implies r = 4
The coefficient is (54)k4=5k4\binom{5}{4} k^4 = 5k^4.

4. Find the value of r for the term in 1/x

For the term in 1x\frac{1}{x} (or x−1x^{-1}):
5−2r=−15 - 2r = -1
2r=6  ⟹  r=32r = 6 \implies r = 3
The coefficient is (53)k3=10k3\binom{5}{3} k^3 = 10k^3.

5. Equate the coefficients and solve for k

Since the coefficients are the same:
5k4=10k35k^4 = 10k^3
Since kk is a positive integer, k≠0k \neq 0, so we can divide by 5k35k^3:
k=2k = 2

答案

k=2k = 2

6(b)

1. Find the general term for the expansion of $(x + \frac{k}{x})^5$.

The general term is given by:
Tr+1=(5r)x5−r(kx)r=(5r)krx5−2rT_{r+1} = \binom{5}{r} x^{5-r} \left(\frac{k}{x}\right)^r = \binom{5}{r} k^r x^{5-2r}

2. Identify the powers of $x$ required from the binomial expansion to form a constant term.

To obtain a constant term (i.e., x0x^0) in the expansion of (1−3x2)(x+kx)5\left(1 - 3x^2\right)\left(x + \frac{k}{x}\right)^5, we need to multiply:

  • 11 by the x0x^0 term from (x+kx)5\left(x + \frac{k}{x}\right)^5
  • −3x2-3x^2 by the x−2x^{-2} term from (x+kx)5\left(x + \frac{k}{x}\right)^5

3. Show that these required powers of $x$ do not exist in the binomial expansion.

For the x0x^0 term: 5−2r=0  ⟹  r=2.55 - 2r = 0 \implies r = 2.5
For the x−2x^{-2} term: 5−2r=−2  ⟹  r=3.55 - 2r = -2 \implies r = 3.5
Since rr must be an integer (0⩽r⩽50 \leqslant r \leqslant 5), neither the x0x^0 term nor the x−2x^{-2} term exists in the expansion of (x+kx)5\left(x + \frac{k}{x}\right)^5. Therefore, there is no constant term in the overall expansion.

答案

Since rr must be an integer, there are no x0x^0 or x−2x^{-2} terms in the expansion of (x+kx)5\left(x + \frac{k}{x}\right)^5, hence no constant term can be formed.

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