1、课时规范练37数学归纳法基础巩固组1.用数学归纳法证明不等式2n(n+1)2(nN+)时,初始值n应等于()A.1B.4C.5D.62.(2020江西高二期中)用数学归纳法证明不等式1n+1+1n+2+1n+3+12n45-12n+1(n2,nN+)的过程中由n=k递推到n=k+1时不等式左边应添加的项为()A.12k+2B.12k+1+12k+2C.12k+1+12k+2-1k+1D.12k+1+12k+2-1k+1-1k+23.已知nN+,用数学归纳法证明f(n)=1+4+7+(3n-2)=3n2-n2时.假设当n=k(kN+)时命题成立,证明当n=k+1时命题也成立,需要用到的f(k+1
2、)与f(k)之间的关系式是()A.f(k+1)=f(k)+3k-5B.f(k+1)=f(k)+3k-2C.f(k+1)=f(k)+3k+1D.f(k+1)=f(k)+3k+44.(2020河北宣化一中月考)用数学归纳法证明1+2+3+n2=n4+n22,则当n=k+1时左端应在n=k的基础上()A.增加一项B.增加2k项C.增加2k项D.增加2k+1项5.(2020浙江高三期末)对于不等式n2+2nn+2(nN+),某同学用数学归纳法证明的过程如下:当n=1时,12+21+2,不等式成立;假设当n=k(nN+)时,不等式成立,即k2+2kk+2,则当n=k+1时,(k+1)2+2(k+1)=k
3、2+4k+30,所以a1=1,当n=2时,13+23=a2322,解得a2=2,当n=3时,13+23+33=a3422,解得a3=3.(2)猜想数列an的通项公式为an=n,当n=1时,由(1)可知结论成立;假设当n=k时,结论成立,即ak=k成立,则n=k+1时,由13+23+k3=ak(k+1)22与13+23+(k+1)3=ak+1(k+2)22,所以(k+1)3=ak+1(k+2)22-ak(k+1)22=ak+1(k+2)22-k(k+1)22,所以ak+12(k+2)2=4(k+1)3+k2(k+1)2=(k+1)2(4k+4+k2)=(k+1)2(k+2)2.又an0,ak+1
4、=k+1成立,根据猜想成立.13.Cf(n)=1+12+13+13n-1+13n(nN*),P=f(k)=1+12+13+13k-1+13k(kN*),P+Q=f(k+1)=1+12+13+13k-1+13k+13k+1+13k+2+13k+1-1+13k+1.Q=f(k+1)-f(k)=13k+1+13k+2+13k+1-1+13k+1.故选C.14.(1)解32=34=12.-32=-3(-3+2-1)=6.(2)证明利用数学归纳法证明.r=1时,a+b1=C10a1b0+C11a0b1=a+b成立.假设r=n时成立,则a+bn=k=0nCnkan-kbk成立,则r=n+1时,a+bn+1=a+bna+b1=(Cn0an+Cn1an-1b1+Cnnbn)a+b=(Cn0an+1+Cn1anb1+Cnnabn)+(Cn0anb+Cn1an-1b2+Cnnbn+1).根据Cnk+Cnk+1=Cn+1k+1.a+bn+1=Cn0an+1+Cn+11anb1+Cn+1n+1bn+1=k=0n+1Cn+1kan+1-kbk.假设成立,即r=n+1时命题成立.综上可得,a+br=k=0rCrkar-kbk,a,bR,rN*.
