《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]
