设数列{an}的前n项和为Sn,且Sn=n2+2n
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)证明{an}是等差数列.
解:(Ⅰ)∵Sn=n2+2n,
∴a1=S1=3,
当n≥2时,an=Sn﹣Sn﹣1=n2+2n﹣(n﹣1)2+2(n﹣1)=2n+1,
则当n=1时,满足an=2n+1,综上都有an=2n+1.
读我国主要山脉分布图,完成下列各题.
点P(1,3)在直线l:x﹣2y+1=0的( )
已知a1,a2,a3,a4是各项均为正数的等差数列,其公差d大于零,若线段l1,l2,l3,l4的长分别为a1,a2,a3,a4,则( )
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