【導(dǎo)讀】解析:∵an=1n+2,∴a10=D.解析:由已知得a2=2×2-2==3×3+1=10,4.?dāng)?shù)列{an}滿足an+an+1=12,a2=1,Sn是數(shù)列{an}的前n項(xiàng)和,則S21為(). 解析:由已知可得a1=-12,a3=-12,a4=1,?S21=S20+a21=102-12=92.解析:由a1=1=4-3,a2=5=8-3,a3=13=16-3,a4=29=32-3,猜測an=2n+1. 故第n個(gè)圖中點(diǎn)的個(gè)數(shù)為(n-1)×n+1=n2-n+1.a5=3a4-2a3=3×15-2×7=31.令2n-1=127,得2n=27,解得n=7.5350是否是這個(gè)數(shù)列中的項(xiàng)?若有,求出該項(xiàng);若沒有,說明理由.。解析:∵an+1-an=-3<0,由遞減數(shù)列的定義知B選項(xiàng)正確.故選B.∵n∈N*,∴an+1-an<C.從而an=7t2-3t=72-928.函數(shù)f=7t2-3t在(0,314]上是減函數(shù),在[314,1]上是增函數(shù),所以a1是最大項(xiàng),故。S11=1-5+9-13+?+33-37+41=21,或S22-S11=a12+a13+?+a22=a12+++?,發(fā)現(xiàn)周期為6,則a2021=a3