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<p><span lang=EN-US> x+y=2k+2时，有k种方案，即“2，2k”、“3，2k-1”、...…、“k+1，k+1”。 </span></p>
<p><span lang=EN-US> x+y=2k+3时，有k种方案，即“3，2k”、“4，2k”、...…、“k+2，k+2”。 </span></p>
<p><span lang=EN-US> … … </span></p>
<p><span lang=EN-US> … … </span></p>
<p><span lang=EN-US> x+y=4k-1时，有1种方案，即“2k-1，2k”。 </span></p>
<p><span lang=EN-US> x+y=4k时，有1种方案，即“2k，2k”。 </span></p>
<p><span lang=EN-US> <img width=137
height=112 id="_x0000_i1142" src="timu.files\timu.h25.gif" border=0></span></p>
<p>（<span lang=EN-US>b） <img width=167 height=69 id="_x0000_i1143"
src="timu.files\timu.h26.gif" border=0></span></p>
<p><span lang=EN-US> <img width=142
height=38 id="_x0000_i1144" src="timu.files\timu.h27.gif" border=0></span></p>
<p><span lang=EN-US> <img width=224
height=115 id="_x0000_i1145" src="timu.files\timu.h28.gif" border=0></span></p>
<p><span lang=EN-US> <img width=194
height=101 id="_x0000_i1146" src="timu.files\timu.h29.gif" border=0></span></p>
<p><span lang=EN-US> <img width=180
height=104 id="_x0000_i1147" src="timu.files\timu.h30.gif" border=0></span></p>
<p><span lang=EN-US> <img width=216
height=124 id="_x0000_i1148" src="timu.files\timu.h31.gif" border=0></span></p>
<p><span lang=EN-US> <img width=151
height=83 id="_x0000_i1149" src="timu.files\timu.h32.gif" border=0></span></p>
<p><span lang=EN-US> </span></p>
<p><span lang=EN-US>27. 设 <span style='mso-text-raise:-3.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1542" type="#_x0000_t75" style='width:27pt;height:13.8pt' o:ole="">
<v:imagedata src="./timu.files/image074.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=36 height=18
src="./timu.files/image075.gif" v:shapes="_x0000_i1542"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1542"
DrawAspect="Content" ObjectID="_1069585722">
</o:OLEObject>
</xml><![endif]-->,<span style='mso-text-raise:-15.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1547" type="#_x0000_t75" style='width:78pt;height:36pt' o:ole="">
<v:imagedata src="./timu.files/image076.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=104 height=48
src="./timu.files/image077.gif" v:shapes="_x0000_i1547"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1547"
DrawAspect="Content" ObjectID="_1069585723">
</o:OLEObject>
</xml><![endif]-->,<span style='mso-text-raise:-15.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1552" type="#_x0000_t75" style='width:81pt;height:36pt' o:ole="">
<v:imagedata src="./timu.files/image078.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=108 height=48
src="./timu.files/image079.gif" v:shapes="_x0000_i1552"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1552"
DrawAspect="Content" ObjectID="_1069585724">
</o:OLEObject>
</xml><![endif]--></span></p>
<p>（<span lang=EN-US>a）证明 a<sub>n+1</sub>=a<sub>n</sub>+b<sub>n+1</sub>, b<sub>n+1</sub>=a<sub>n</sub>+b<sub>n</sub></span></p>
<p>（<span lang=EN-US>b）求序列{a<sub>n</sub>}与{b<sub>n</sub>}的母函数。 </span></p>
<p>（<span lang=EN-US>c）用Fibonacci数来表示a<sub>n</sub>与b<sub>n</sub>。 </span></p>
<p>解：<span lang=EN-US>1）证明 </span></p>
<p><span lang=EN-US> <img width=240
height=104 id="_x0000_i1151" src="timu.files\timu.h33.gif" border=0></span></p>
<p><span lang=EN-US> <img width=230
height=83 id="_x0000_i1152" src="timu.files\timu.h34.gif" border=0></span></p>
<p>同理可证<span lang=EN-US> <img width=72 height=20 id="_x0000_i1153"
src="timu.files\timu.h35.gif" border=0></span></p>
<p><span lang=EN-US>2)求解方程组：<span style='mso-text-raise:-15.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1559" type="#_x0000_t75" style='width:124.2pt;height:36pt' o:ole="">
<v:imagedata src="./timu.files/image080.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=166 height=48
src="./timu.files/image081.gif" v:shapes="_x0000_i1559"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1559"
DrawAspect="Content" ObjectID="_1069585725">
</o:OLEObject>
</xml><![endif]-->，知道<span style='mso-text-raise:-12.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1564" type="#_x0000_t75" style='width:91.8pt;height:31.2pt' o:ole="">
<v:imagedata src="./timu.files/image082.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=122 height=42
src="./timu.files/image083.gif" v:shapes="_x0000_i1564"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1564"
DrawAspect="Content" ObjectID="_1069585726">
</o:OLEObject>
</xml><![endif]-->，<span style='mso-text-raise:-12.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1569" type="#_x0000_t75" style='width:90pt;height:31.2pt' o:ole="">
<v:imagedata src="./timu.files/image084.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=120 height=42
src="./timu.files/image085.gif" v:shapes="_x0000_i1569"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1569"
DrawAspect="Content" ObjectID="_1069585727">
</o:OLEObject>
</xml><![endif]--></span></p>
<p><span lang=EN-US>3）解 <img width=120 height=41
id="_x0000_i1154" src="timu.files\timu.h36.gif" border=0></span></p>
<p><span lang=EN-US>28. 设 F<sub>1</sub>=F<sub>2</sub>=1, F<sub>n</sub>=F<sub>n-1</sub>+F<sub>n-2</sub></span></p>
<p>（<span lang=EN-US>a）证明 <span style='mso-text-raise:-6.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1576" type="#_x0000_t75" style='width:121.2pt;height:18pt' o:ole="">
<v:imagedata src="./timu.files/image086.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=162 height=24
src="./timu.files/image087.gif" v:shapes="_x0000_i1576"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1576"
DrawAspect="Content" ObjectID="_1069585728">
</o:OLEObject>
</xml><![endif]-->，n > k > 1.</span></p>
<p>（<span lang=EN-US>b）证明F<sub>m</sub>|F<sub>n</sub>的充要条件是m|n。 </span></p>
<p>（<span lang=EN-US>c）证明 <span style='mso-text-raise:-16.0pt'><!--[if gte vml 1]><v:shape
id="_x0000_i1588" type="#_x0000_t75" style='width:285pt;height:37.8pt' o:ole="">
<v:imagedata src="./timu.files/image088.wmz" o:title=""/>
</v:shape><![endif]--><![if !vml]><img width=380 height=50
src="./timu.files/image089.gif" v:shapes="_x0000_i1588"><![endif]></span><!--[if gte mso 9]><xml>
<o:OLEObject Type="Embed" ProgID="Equation.3" ShapeID="_x0000_i1588"
DrawAspect="Content" ObjectID="_1069585729">
</o:OLEObject>
</xml><![endif]--></span></p>
<p>（<span lang=EN-US>d）证明(F<sub>m</sub>,F<sub>n</sub>)=F<sub>(m,n)</sub>, (m,n)为m,n的最大公约数。 </span></p>
<p>解：<span lang=EN-US>1）证明: 用数学归纳法 </span></p>
<p><span lang=EN-US> I k=2时 成立，即 <img width=110 height=20
id="_x0000_i1157" src="timu.files\timu.h37.gif" border=0></span></p>
<p><span lang=EN-US> II 设k=m时成立 <img width=134 height=20
id="_x0000_i1158" src="timu.files\timu.h38.gif" border=0></span></p>
<p><span lang=EN-US> 则k=m+1时， </span></p>
<p><span lang=EN-US>
<img width=203 height=104 id="_x0000_i1159" src="timu.files\timu.h39.gif"
border=0></span></p>
<p><span lang=EN-US> 用归纳假设 由I、II知题设成立。 </span></p>
<p><span lang=EN-US> 2）证明 : F<sub>m</sub>与F<sub>m-1</sub>互素（自己证明） </span></p>
<p><span lang=EN-US> <img width=187
height=81 id="_x0000_i1160" src="timu.files\timu.h40.gif" border=0></span></p>
<p>作了<span lang=EN-US>k次后 若n-km<m</span></p>
<p>则上式</p>
<p><span lang=EN-US> <img width=77 height=38 id="_x0000_i1161"
src="timu.files\timu.h41.gif" border=0></span></p>
<p><span lang=EN-US>3）证明 </span></p>
<p><span lang=EN-US> <img width=159
height=101 id="_x0000_i1162" src="timu.files\timu.h42.gif" border=0></span></p>
<p>当<span lang=EN-US>n是偶数时，最后一次会出现F<sub>0</sub>=0项 </span></p>
<p><span lang=EN-US> <img width=194
height=40 id="_x0000_i1163" src="timu.files\timu.h43.gif" border=0></span></p>
<p>当<span lang=EN-US>n是奇数时，最后一次会出现F<sub>1</sub>=1项 </span></p>
<p><span lang=EN-US> <img width=201
height=41 id="_x0000_i1164" src="timu.files\timu.h44.gif" border=0></span></p>
<p><span lang=EN-US>4）证明 : 用2）的结论 </span></p>
<p><span lang=EN-US> <img width=134
height=41 id="_x0000_i1165" src="timu.files\timu.h45.gif" border=0></span></p>
<p>下面证明是最大公约数<span lang=EN-US> </span></p>
<p>设<span lang=EN-US>F<sub>(n,m)</sub>不是最大公约数， F<sub>N</sub>是,</span></p>
<p>则<span lang=EN-US> m|N, n|N 则 N=<(n,m)与 N>(n,m)矛盾 </span></p>
<p><span lang=EN-US>29. 从1到n的自然数中选取k个不同且不相邻的数，设此选取的方案为f(n,k)。 </span></p>
<p>（<span lang=EN-US>a）求f(n,k)的递推关系。 </span></p>
<p>（<span lang=EN-US>b）用归纳法求f(n,k)。 </span></p>
<p>（<span lang=EN-US>c）若设1与n算是相邻的数，并设在此假定下从1到n的自然数中选取k个不同且不相邻的k个数的方案数为g(n,k)，利用f(n,k)求g(n,k)。 </span></p>
<p>解：（<span lang=EN-US>a） </span></p>
<p><span lang=EN-US> <img width=195
height=61 id="_x0000_i1166" src="timu.files\timu.h46.gif" border=0></span></p>
<p>（<span lang=EN-US>b） </span></p>
<p><span lang=EN-US> <img width=195
height=61 id="_x0000_i1167" src="timu.files\timu.h47.gif" border=0></span></p>
<p>（<span lang=EN-US>c）</span></p>
<p><span lang=EN-US> <img width=181
height=124 id="_x0000_i1168" src="timu.files\timu.h48.gif" border=0></span></p>
<p><span lang=EN-US>30. 设S<sub>2</sub>(n,k)是第二类Stirling数。</span></p>
<p>证明<span lang=EN-US> :设与第n+1号球同盒的球有n-k个，这样，其他k个球就放入另外m-1个盒子， k=m-1,m,…,n。
即从n个不同的球中取k个放入m-1个相同的盒子的方案有 </span></p>
<p><span lang=EN-US> <img width=211
height=126 id="_x0000_i1170" src="timu.files\timu.h49.gif" border=0></span></p>
<p><span lang=EN-US>31. 求下图中从A点出发到n点的路径数。 </span></p>
<p>解：把上方的点序列设为<span lang=EN-US>a<sub>n</sub> ,<sub> </sub>把下方的点序列设为b<sub>n</sub></span></p>
<p><span lang=EN-US> <sub> </sub>可得递推关系 </span></p>
<p><span lang=EN-US> <img width=201
height=63 id="_x0000_i1172" src="timu.files\timu.h50.gif" border=0></span></p>
<p>特征方程为 <span lang=EN-US><img width=79 height=18 id="_x0000_i1173"
src="timu.files\timu.h51.gif" border=0></span></p>
<p>解得<span lang=EN-US> <img width=61 height=39
id="_x0000_i1174" src="timu.files\timu.h52.gif" border=0></span></p>
<p>代入初值可解。</p>
<p><span lang=EN-US>32. n位0,1符号串，求从左向右只在最后两位才出现0，0的符号串的数目。</span></p>
<p>解：设所求的串的个数为<s⌨️ 快捷键说明
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