湘教版高中數(shù)學(xué)選擇性必修第一冊(cè)第1章數(shù)列１-４數(shù)學(xué)歸納法練習(xí)含答案

*1.4數(shù)學(xué)歸納法基礎(chǔ)過(guò)關(guān)練題組一利用數(shù)學(xué)歸納法證明等式1.(2022湖南湘潭一中月考)已知n為正偶數(shù),用數(shù)學(xué)歸納法證明1-12+13-14+…+1n-1-1n=21n+2+1A.n=k+1時(shí)等式成立B.n=k+2時(shí)等式成立C.n=2k+2時(shí)等式成立D.n=2(k+2)時(shí)等式成立2.(2021安徽合肥肥東二中月考)用數(shù)學(xué)歸納法證明等式1+2+3+…+(2n+1)=(n+1)(2n+1)(n∈N+)時(shí),從n=k到n=k+1,等式左邊需增添的項(xiàng)是()A.2k+2B.2(k+1)+1C.(2k+2)+(2k+3)D.[(k+1)+1][2(k+1)+1]3.用數(shù)學(xué)歸納法證明1-141-191-116×…×1-1題組二利用數(shù)學(xué)歸納法證明不等式4.用數(shù)學(xué)歸納法證明1n+1+1n+2+…+13n+1>1(n∈N+A.12+13+14B.C.125.對(duì)于不等式n2+n<n+1(n∈N+(1)當(dāng)n=1時(shí),12+1<1+1,(2)假設(shè)當(dāng)n=k(k≥1且k∈N+)時(shí),不等式成立,即k2則當(dāng)n=k+1時(shí),(=k2+3=(k所以當(dāng)n=k+1時(shí),不等式成立.則上述證法()A.過(guò)程全部正確B.n=1驗(yàn)證不正確C.歸納假設(shè)不正確D.從n=k到n=k+1的推理不正確6.用數(shù)學(xué)歸納法證明“1+12+13+…+12n-1<n(n∈N+,n>1)”的過(guò)程中,從n=k(k∈N+,k>1)到A.kB.2kC.2k-1D.2k-17.用數(shù)學(xué)歸納法證明:112+132+…+1(2n-1)2>1-12+13-1題組三歸納—猜想—證明解決與遞推公式有關(guān)的數(shù)列問(wèn)題8.(2021江西上饒?jiān)驴?已知數(shù)列{an}的前n項(xiàng)和為Sn,a2=14,且an=12+1nSn-2n-1(n(1)求S12,S2(2)由(1)猜想數(shù)列Sn2n的通項(xiàng)公式9.(2021陜西西安鐵一中學(xué)期末)在數(shù)列{an}中,a1=12,an+1=3(1)求出a2,a3,并猜想{an}的通項(xiàng)公式;(2)用數(shù)學(xué)歸納法證明(1)中的猜想.

答案與分層梯度式解析基礎(chǔ)過(guò)關(guān)練1.B因?yàn)閚為正偶數(shù),所以當(dāng)n=k時(shí),下一個(gè)偶數(shù)為k+2.2.C當(dāng)n=k(k∈N+,k>1)時(shí),左邊=1+2+3+…+(2k+1),當(dāng)n=k+1時(shí),左邊=1+2+3+…+(2k+1)+(2k+2)+(2k+3),故選C.3.證明(1)當(dāng)n=2時(shí),左邊=1-14=34,右邊=2+12×2=3(2)假設(shè)當(dāng)n=k(k≥2,k∈N+)時(shí),等式成立,即1-141-191-1那么當(dāng)n=k+1時(shí),1-141-191-116×…×1-1k21-1(k+1)綜合(1)(2)知,對(duì)任意n≥2,n∈N+等式恒成立.4.A5.D在n=k+1時(shí),沒(méi)有應(yīng)用n=k時(shí)的歸納假設(shè),不是數(shù)學(xué)歸納法.6.B由題意知,當(dāng)n=k(k∈N+,k>1)時(shí),左邊=1+12+13+…+12k-1,當(dāng)n=k+1時(shí),左邊=1+12+13+…+12k-1+12k+12k+1+…+17.證明(1)當(dāng)n=1時(shí),左邊=1,右邊=1-12=12,左邊>右邊,(2)假設(shè)當(dāng)n=k(k∈N+)時(shí),不等式成立,即112+132+…+1(2k-1)2>1-12+則當(dāng)n=k+1時(shí),112+132+…>1-12+13-14+…+12>1-12+13-14+…+12=1-12+13-14+…+12k-1-即當(dāng)n=k+1時(shí),不等式也成立.由(1)(2)知,不等式對(duì)任何n∈N+都成立.8.解析(1)∵an=12+1nSn-2n-1(n∴當(dāng)n=1時(shí),a1=S1=12+1S1-1,解得S1=2,則當(dāng)n=2時(shí),a2=12+12S2-2=14,解得S2當(dāng)n=3時(shí),a3=S3-S2=12+13S3-22,解得S3(2)由(1)猜想數(shù)列Sn2n的通項(xiàng)公式為Sn2n=n用數(shù)學(xué)歸納法證明如下:①當(dāng)n=1時(shí),由(1)可得S12=1,當(dāng)n=2時(shí),由(1)可得S24=4,②假設(shè)當(dāng)n=k(k∈N+,k≥2)時(shí),Sk2k則當(dāng)n=k+1時(shí),ak+1=Sk+1-Sk=12+1k+1即12-1k+1Sk+1=Sk-2k=2k·k2-2k=(k2則k-12(k+1)Sk+1=(k+1)(k-1)因?yàn)閗≥2,所以Sk+1=2(k+1)2·2k=(k+1)2·2k+1,即Sk+12所以當(dāng)n=k+1時(shí),結(jié)論也成立.由①②可知,對(duì)任何n∈N+,Sn2n=n9.解析(1)∵a1=12,an+1=3∴a2=3a1a1+3=3×1212+3=猜想:an=3n+5(n∈N(2)證明:①