{an}是首项a1=-10,公差d1=2的等差数列,{bn}是首项b1=-1/2,公差d2=1/2的等差数列,向量OA=

学习 时间:2026-04-07 18:11:39 阅读:3530
{an}是首项a1=-10,公差d1=2的等差数列,{bn}是首项b1=-1/2,公差d2=1/2的等差数列,向量OA=(-1,1)

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孤独的柠檬

靓丽的奇迹

2026-04-07 18:11:39

为了避免用数学归纳法,假设:3≤k≤n,即k是3至n中的任意自然数则:OP(k)-OP(2)=akOA+bkOB-a2OA-b2OB=(ak-a2)OA+(bk-b2)OB=(k-2)d1OA+(k-2)d2OB=2(k-2)OA+(1/2)(k-2)OB=(k-2)(2OA+OB/2)=(k-2)(-3/2,5/2)OP(k)-OP(1)=akOA+bkOB-a1OA-b1OB=(ak-a1)OA+(bk-b1)OB=(k-1)d1OA+(k-1)d2OB=2(k-1)OA+(1/2)(k-1)OB=(k-1)(2OA+OB/2)=(k-1)(-3/2,5/2)可以看出,对于任意给定的满足3≤k≤n的k值,OP(k)-OP(2)=((k-2)/(k-1))(OP(k)-OP(1))即:P1,P2,。。。,Pn共线OP(k)=akOA+bkOB=(a1+(k-1)d1)OA+(b1+(k-1)d2)OB=(2k-12)OA+(k/2-1)OB=(2k-12)(-1,1)+(k/2-1)(1,1)=(11-3k/2,5k/2-13)即:11-3k/2>0,5k/2-13>0,即:k26/5,即:k=6或7

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  • 潇洒的故事
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    2026-04-07 18:11:39

    为了避免用数学归纳法,假设:3≤k≤n,即k是3至n中的任意自然数则:OP(k)-OP(2)=akOA+bkOB-a2OA-b2OB=(ak-a2)OA+(bk-b2)OB=(k-2)d1OA+(k-2)d2OB=2(k-2)OA+(1/2)(k-2)OB=(k-2)(2OA+OB/2)=(k-2)(-3/2,5/2)OP(k)-OP(1)=akOA+bkOB-a1OA-b1OB=(ak-a1)OA+(bk-b1)OB=(k-1)d1OA+(k-1)d2OB=2(k-1)OA+(1/2)(k-1)OB=(k-1)(2OA+OB/2)=(k-1)(-3/2,5/2)可以看出,对于任意给定的满足3≤k≤n的k值,OP(k)-OP(2)=((k-2)/(k-1))(OP(k)-OP(1))即:P1,P2,。。。,Pn共线OP(k)=akOA+bkOB=(a1+(k-1)d1)OA+(b1+(k-1)d2)OB=(2k-12)OA+(k/2-1)OB=(2k-12)(-1,1)+(k/2-1)(1,1)=(11-3k/2,5k/2-13)即:11-3k/2>0,5k/2-13>0,即:k26/5,即:k=6或7

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