《什么是数学 》习题 第一章 2 数系的无限性 数学归纳法

数学归纳法证明：$1^3 + 2^3 + 3^3 + … + n^3 = [n(n+1)/2]^2$

证明：

如果命题对$n=r$时对情形是正确对，即

$1^3 + 2^3 + 3^3 + … + r^3 = [r(r+1)/2]^2$

然后在这等式两边加上$(r+1)^3$，我们得到

$1^3 + 2^3 + 3^3 + … + r^3+(r+1)^3$

$= [r(r+1)/2]^2 +(r+1)^3$

$= r^2(r+1)^2/2^2 + (r+1)^3$

$= [r^2(r+1)^2+4(r+1)^3]/2^2$

$= (r+1)^2(r^2 +4(r+1))/2^2$

$= (r+1)^2(r+2)^2/2^2$

$= [(r+1)(r+1+1)/2]^2$

这正好是当$n=r+1$时的情形。

而当$n=1$时，

$1^3 = [1(1 + 1)/2]^2= 1$ ，

成立，因此该等式对每个$n$都成立。

数学归纳法证明：$\frac 1 {1*2} + \frac 1 {2*3} + … + \frac 1 {n(n+1)} = \frac n {n+1}$

当$n = 1$时，$\frac 1 {1*2} = \frac 1 {1+1} = \frac 1 2$

当$n = r$时，$\frac 1 {1*2} + \frac 1 {2*3} + … + \frac 1 {r(r+1)} = \frac r {r+1}$

两边加上 $\frac 1 {(r+1)(r+1 + 1)}$

$\frac r {r+1} + \frac 1 {(r+1)(r+1 + 1)}$

$= \frac {r(r+1+1) + 1} {(r+1)(r + 1 + 1)}$

$= \frac {(r+1)^2} {(r+1)(r + 1 + 1)}$

$= \frac {(r+1)} {(r + 1 + 1)}$

这正好是当$n=r+1$时的情形。

数学归纳法证明：$\frac 1 2 + \frac 2 {2^2} + \frac 3 {2^3} + … + \frac n {2^n} = 2 - \frac {n+2} {2^n}$

$2 - \frac {n+2} {2^n} + \frac {n+1} {2^{n+1}}$

$=2 - \frac {2(n+2)-(n+1)} {2^{n+1}}$

$=2 - \frac {n+1+2} {2^{n+1}}$

数学归纳法证明：$1 + 2q + 3q^2 + … + nq^{n-1} = \frac {1-(n+1)q^n+nq^{n+1}} {(1-q)^2}$

$\frac {1-(n+1)q^n + nq^{n+1}} {(1-q)^2} + (n+1)q^n$

$= \frac {1-(n+1)q^n + nq^{n+1} + (n+1)q^n(1-q)^2} {(1-q)^2}$

$= \frac {1-(n+1)q^n + nq^{n+1} + (n+1)q^n(q^2-2q+1)} {(1-q)^2}$

$= \frac {1 + nq^{n+1} + (n+1)q^n(q^2-2q)} {(1-q)^2}$

$= \frac {1 + nq^{n+1} + (n+1)q^{n+2} -2(n+1)q^{n+1}} {(1-q)^2}$

$= \frac {1 -(n+2)q^{n+1} + (n+1)q^{n+2} } {(1-q)^2}$

数学归纳法证明：$(1+q)(1+q^2)(1+q^4)…(1+q^{2^n}) = \frac {1-q^{2^{n+1}}} {1-q}$

$(\frac {1-q^{2^{n+1}}} {1-q})(1+q^{2^{n+1}})$

$= \frac {1-q^{2^{n+1+1}}} {1-q}$

求出下列等比级数的和：

$\frac 1 {1+x^2} + \frac 1 {(1+x^2)^2} + … + \frac 1 {(1+x^2)^n}$

设 ：

$\frac 1 {1+x^2} + \frac 1 {(1+x^2)^2} + … + \frac 1 {(1+x^2)^n} = m$ (1)

$(\frac 1 {1+x^2} + \frac 1 {(1+x^2)^2} + … + \frac 1 {(1+x^2)^n}) * (1+x^2) = m*(1+x^2)$

$1 + \frac 1 {1+x^2} + … + \frac 1 {(1+x^2)^{n-1}} = m*(1+x^2)$ (2)

(2) - (1) 得

$1 - \frac 1 {(1+x^2)^n} = mx^2$

$m = \frac {1- \frac 1 {(1+x^2)^n} } {x^2}$

$1 + \frac x {1+x^2} + \frac {x^2} {(1+x^2)^2} + … + \frac {x^n} {(1+x^2)^n}$

设 ：

$1 + \frac x {1+x^2} + \frac {x^2} {(1+x^2)^2} + … + \frac {x^n} {(1+x^2)^n} = m$ (1)

$\frac x {1+x^2} + \frac {x^2} {(1+x^2)^2} + … + \frac {x^{n+1}} {(1+x^2)^{n+1}} = m * \frac x {1+x^2}$ (2)

(2) - (1) 得

$\frac {x^{n+1}} {(1+x^2)^{n+1}} - 1 = m * \frac x {1+x^2} - m$

$\frac {x^{n+1} - (1+x^2)^{n+1}} {(1+x^2)^{n+1}} = m * \frac {x - (1+x^2)} {1+x^2}$

$\frac {x^{n+1} - (1+x^2)^{n+1}} {(1+x^2)^n} = m * [x - (1+x^2)]$

$m = \frac {x^{n+1} - (1+x^2)^{n+1}} {x(1+x^2)^n - (1+x^2)(1+x^2)^n}$

$\frac {x^2 - y^2} {x^2 + y^2} + (\frac {x^2 - y^2} {x^2 + y^2})^2 + … + (\frac {x^2 - y^2} {x^2 + y^2})^n$

设 ：

$\frac {x^2 - y^2} {x^2 + y^2} + (\frac {x^2 - y^2} {x^2 + y^2})^2 + … + (\frac {x^2 - y^2} {x^2 + y^2})^n = m$ (1)

$(\frac {x^2 - y^2} {x^2 + y^2})^2 + (\frac {x^2 - y^2} {x^2 + y^2})^3 + … + (\frac {x^2 - y^2} {x^2 + y^2})^{n+1} = m(\frac {x^2 - y^2} {x^2 + y^2})$ (2)

(2) - (1) 得

$(\frac {x^2 - y^2} {x^2 + y^2})^{n+1} - \frac {x^2 - y^2} {x^2 + y^2} = m(\frac {x^2 - y^2} {x^2 + y^2}) - m$

$(\frac {x^2 - y^2} {x^2 + y^2})^{n+1} - \frac {x^2 - y^2} {x^2 + y^2} = m\frac {x^2 - y^2 - (x^2 + y^2)} {x^2 + y^2}$

不算了恶心了，反正等比级数求和公式如下

$S_n = \frac {a_1(1-q^n)} {1-q}$

用公式（4）和（5）证明：

$1^2 + 3^2 +…+ (2n + 1)^2 = \frac {(n+1)(2n+1)(2n+3)} {3}$

已知：$1^2 + 2^2 +…+ n^2 = \frac {n(n+1)(2n+1)} {6}$

$2^2 + 4^2 +…+ (2n)^2$

$= 2^2 * 1^2 + 2^2*2^2 +…+ 2^2*n^2$

$= 2^2(1^2 + 2^2 +…+ n^2)$

$=2^2 \frac {n(n+1)(2n+1)} {6}$

$1^2 + 2^2 +…+ (2n+1)^2 = \frac {(2n+1)(2n+1+1)(4n+2+1)} {6}$

$1^2 + 3^2 +…+ (2n + 1)^2 = [1^2 + 2^2 +…+ (2n+1)^2] - [2^2 + 4^2 +…+ (2n)^2]$

$= \frac {(2n+1)(2n+1+1)(4n+2+1)} {6} - 2^2 \frac {n(n+1)(2n+1)} {6}$

$= \frac {(2n+1)(n+1)(4n+3)} {3} - \frac {2n(n+1)(2n+1)} {3}$

$= \frac {(n+1)(2n+1)(2n+3)} {3}$

$1^3 + 3^3 +…+ (2n + 1)^3 = (n + 1)^2(2n^2 + 4n + 1)$

已知：$1^3 + 2^3 +…+ n^3 = [\frac {n(n+1)}{2}]^2$

$2^3 + 4^3 +…+ (2n)^3$

$= 2^3 * 1^3 + 2^3*2^3 +…+ 2^3*n^3$

$= 2^3(1^3 + 2^3 +…+ n^3)$

$= 2^3[\frac {n(n+1)}{2}]^2$

$1^3 + 2^3 +…+ (2n+1)^3 = [\frac {(2n + 1)(2n + 1 + 1)}{2}]^2$

$1^3 + 3^3 +…+ (2n + 1)^3 = [1^3 + 2^3 +…+ (2n+1)^3] - [2^3 + 4^3 +…+ (2n)^3]$

$= [\frac {(2n + 1)(2n + 1 + 1)}{2}]^2 - 2^3[\frac {n(n+1)}{2}]^2$

$= (n+1)^2(2n+1)^2 - 2n^2(n+1)^2$

$= (n + 1)^2(2n^2 + 4n + 1)$

数学归纳法证明：

$1^2 + 3^2 +…+ (2n + 1)^2 = \frac {(n+1)(2n+1)(2n+3)} {3}$

$1^2 + 3^2 +…+ (2n + 3)^2 = \frac {(n+1)(2n+1)(2n+3)} {3} + (2n + 3)^2$

$= \frac {(2n+3)(2n+5)(n+2)} {3}$

$= \frac {(n+ 1 + 1)(2n + 2 + 1)(2n + 2 + 3)} {3}$

$1^3 + 3^3 +…+ (2n + 1)^3 = (n + 1)^2(2n^2 + 4n + 1)$

$1^3 + 3^3 +…+ (2n + 3)^3 = (n + 1)^2(2n^2 + 4n + 1) + (2n + 3)^3$

$= 2n^4 + 4n^3 + n^2 + 4n^3 + 8n^2 + 2n + 2n^2 + 4n + 1 + 8n^3 + 3 * 3 * 4 n^2 + 3 * 2n * 9 + 27$

$= 2n^4 + 16n^3 + 47n^2 + 60n + 28$

$= (n^2 + 4n + 4)(2n^2 + 8n + 7)$

$= (n + 2)^2[2(n + 1)^2 + 4(n + 1) + 1]$