单元测评卷(三)1
C[解析]由条件得a2=3,a3=7,a4=15,a5=31,a6=63,故选C
C[解析]由a1+a5=10,及等差数列的性质知2a3=10,即a3=5,∴S6=a1+a62×6=a3+a42×6=5+72×6=36,故选C
BC[解析]取a=-2,b=1,则1a>1b不成立,故A不正确
若00,∴b+1a+1>ba,故C正确;若c0,∴q=2,∴a1=132
a1+a2+a3+…+an=132×(1-2n)1-2=2n-132,∴2n-132>132n×2n(n-1)2,整理可得2n-1>2(n-1)(12n-5)
令n>(n-1)(12n-5)且n≠1,得10,所以a1=9,所以Sn=9n+n(n-1)2×(-2)=-(n-5)2+25,所以Sn的最大值为S5=25
A[解析]由题意得n+1an+1=2an2+4nan+n2an2=(nan)2+4·nan+2,∴n+1an+1+2=(nan+2)2
令bn=nan+2,则bn+1=bn2,两边取对数得lgbn+1=2lgbn
又lgb1=lg(1a1+2)=lg3,∴数列{lgbn}是首项为lg3,公比为2的等比数列,∴lgbn=2n-1·lg3=lg32n-1,∴bn=32n-1,即nan+2=32n-1,∴an=n32n-1-2,∴a7=7326-2,又a7=73λ-2,∴λ=26=64,故选A
4[解析]设{an}的公比为q,由a5+a7=4(a1+a3),得q4=4,∴a6a2=q4=4
(-8,0][解析]由题知,对于一切实数x,mx2-mx-2
