练习题
数列 \( u_1, u_2, \ldots, u_n, \ldots \) 由公式 \( u_n = 3\left(\frac{2}{3}\right)^n - 1 \) 给出，其中 n 是正整数。
a) 求 \( u_1, u_2 \) 和 \( u_3 \) 的值
b) 证明 \( \sum_{n=1}^{15} u_n = -9.014 \) (精确到4位有效数字)
c) 证明 \( u_{n+1} = \frac{2u_n - 1}{3} \)
a) \( u_1 = 3\left(\frac{2}{3}\right)^1 - 1 = 2 - 1 = 1 \)
\( u_2 = 3\left(\frac{2}{3}\right)^2 - 1 = 3 \times \frac{4}{9} - 1 = \frac{4}{3} - 1 = \frac{1}{3} \)
\( u_3 = 3\left(\frac{2}{3}\right)^3 - 1 = 3 \times \frac{8}{27} - 1 = \frac{8}{9} - 1 = -\frac{1}{9} \)
b) 这是一个几何级数，首项 \( a = 1 \)，公比 \( r = \frac{2}{3} \)
\( S_{15} = \frac{1(1-(\frac{2}{3})^{15})}{1-\frac{2}{3}} = \frac{1-(\frac{2}{3})^{15}}{\frac{1}{3}} = 3(1-(\frac{2}{3})^{15}) \)
\( = 3 - 3(\frac{2}{3})^{15} \approx 3 - 0.014 = 2.986 \)
但题目要求的是 \( \sum_{n=1}^{15} u_n \)，其中 \( u_n = 3\left(\frac{2}{3}\right)^n - 1 \)
所以 \( \sum_{n=1}^{15} u_n = 3\sum_{n=1}^{15}\left(\frac{2}{3}\right)^n - 15 \)
\( = 3 \times \frac{\frac{2}{3}(1-(\frac{2}{3})^{15})}{1-\frac{2}{3}} - 15 \)
\( = 3 \times 2(1-(\frac{2}{3})^{15}) - 15 = 6(1-(\frac{2}{3})^{15}) - 15 \)
\( = 6 - 6(\frac{2}{3})^{15} - 15 = -9 - 6(\frac{2}{3})^{15} \approx -9.014 \)
c) \( u_{n+1} = 3\left(\frac{2}{3}\right)^{n+1} - 1 = 3 \times \frac{2}{3} \times \left(\frac{2}{3}\right)^n - 1 \)
\( = 2\left(\frac{2}{3}\right)^n - 1 \)
\( \frac{2u_n - 1}{3} = \frac{2(3(\frac{2}{3})^n - 1) - 1}{3} = \frac{6(\frac{2}{3})^n - 2 - 1}{3} \)
\( = \frac{6(\frac{2}{3})^n - 3}{3} = 2(\frac{2}{3})^n - 1 = u_{n+1} \) ✓
几何级数的第3项和第4项分别是6.4和5.12。求：
a) 级数的公比
b) 级数的首项
c) 级数的无穷和
d) 计算级数的无穷和与前25项和的差
a) \( u_3 = ar^2 = 6.4 \) ... (1)
\( u_4 = ar^3 = 5.12 \) ... (2)
方程(2)除以方程(1)：\( r = \frac{5.12}{6.4} = 0.8 \)
b) 代入方程(1)：\( a(0.8)^2 = 6.4 \)
\( a \times 0.64 = 6.4 \)
\( a = \frac{6.4}{0.64} = 10 \)
c) \( S_{\infty} = \frac{a}{1-r} = \frac{10}{1-0.8} = \frac{10}{0.2} = 50 \)
d) \( S_{25} = \frac{a(1-r^{25})}{1-r} = \frac{10(1-(0.8)^{25})}{0.2} = 50(1-(0.8)^{25}) \)
差 = \( S_{\infty} - S_{25} = 50 - 50(1-(0.8)^{25}) = 50(0.8)^{25} \)
汽车价格每年贬值15%。新车价格为$20,000。
a) 求5年后汽车的价值
b) 求价值低于$4,000的时间
a) 每年贬值15%，意味着每年保留85%的价值
5年后的价值 = \( 20000 \times (0.85)^5 = 20000 \times 0.4437 = \$8,874 \)
b) 设n年后价值低于$4,000
\( 20000 \times (0.85)^n < 4000 \)
\( (0.85)^n < 0.2 \)
取对数：\( n\log(0.85) < \log(0.2) \)
\( n > \frac{\log(0.2)}{\log(0.85)} \approx 9.9 \)
所以10年后价值将低于$4,000
几何级数的前三项是 \( p(3q+1), p(2q+2) \) 和 \( p(2q-1) \)，其中p和q是非零常数。
a) 证明q的一个可能值是5，并求另一个可能值
b) 给定q = 5，且级数的无穷和是896，求前12项的和（精确到2位小数）
a) 在几何级数中，\( \frac{u_2}{u_1} = \frac{u_3}{u_2} \)
\( \frac{p(2q+2)}{p(3q+1)} = \frac{p(2q-1)}{p(2q+2)} \)
\( \frac{2q+2}{3q+1} = \frac{2q-1}{2q+2} \)
\( (2q+2)^2 = (3q+1)(2q-1) \)
\( 4q^2 + 8q + 4 = 6q^2 + 3q - 2q - 1 \)
\( 4q^2 + 8q + 4 = 6q^2 + q - 1 \)
\( 2q^2 - 7q - 5 = 0 \)
\( (2q + 1)(q - 5) = 0 \)
所以 \( q = 5 \) 或 \( q = -\frac{1}{2} \)
b) 当q = 5时：
\( u_1 = p(15+1) = 16p \)
\( u_2 = p(10+2) = 12p \)
\( r = \frac{u_2}{u_1} = \frac{12p}{16p} = \frac{3}{4} \)
\( S_{\infty} = \frac{16p}{1-\frac{3}{4}} = \frac{16p}{\frac{1}{4}} = 64p = 896 \)
所以 \( p = \frac{896}{64} = 14 \)
\( S_{12} = \frac{16 \times 14(1-(\frac{3}{4})^{12})}{1-\frac{3}{4}} = \frac{224(1-(\frac{3}{4})^{12})}{\frac{1}{4}} \)
\( = 896(1-(\frac{3}{4})^{12}) \approx 896(1-0.0317) = 896 \times 0.9683 \approx 867.60 \)
证明等差数列前n项的和是
\( S = \frac{n}{2}[2a + (n-1)d] \)
其中a = 首项，d = 公差
设等差数列为：\( a, a+d, a+2d, \ldots, a+(n-1)d \)
前n项和：\( S_n = a + (a+d) + (a+2d) + \ldots + [a+(n-1)d] \)
倒序排列：\( S_n = [a+(n-1)d] + [a+(n-2)d] + \ldots + (a+d) + a \)
两式相加：\( 2S_n = [2a+(n-1)d] + [2a+(n-1)d] + \ldots + [2a+(n-1)d] \)
\( 2S_n = n[2a+(n-1)d] \)
所以 \( S_n = \frac{n}{2}[2a+(n-1)d] \) ✓
求使 \( \sum_{r=1}^{n}(4r-3) > 2000 \) 的最小n值
\( \sum_{r=1}^{n}(4r-3) = 4\sum_{r=1}^{n}r - 3\sum_{r=1}^{n}1 \)
\( = 4 \times \frac{n(n+1)}{2} - 3n = 2n(n+1) - 3n = 2n^2 + 2n - 3n = 2n^2 - n \)
\( 2n^2 - n > 2000 \)
\( 2n^2 - n - 2000 > 0 \)
解方程 \( 2n^2 - n - 2000 = 0 \)：
\( n = \frac{1 \pm \sqrt{1+16000}}{4} = \frac{1 \pm \sqrt{16001}}{4} \)
\( n \approx \frac{1 \pm 126.5}{4} \)
取正值：\( n \approx \frac{127.5}{4} \approx 31.9 \)
所以最小的n值是32