数列{a
n}的通项为a
n=2n+1,则由b
n=
所确定的数列{b
n}的前n项和是( )
A. n(n+2)
B.
n(n+4)
C.
n(n+5)
D.
n(n+7)
人气:499 ℃ 时间:2020-05-27 11:57:09
解答
∵数列{an}的通项为an=2n+1,∴a1+a2+…+an=2(1+2+…+n)+n=n(n+1)+n,∴bn=a1+a2+…+ann=n(n+1)+nn=n+2,∴数列{bn}的前n项和Sn=(1+2)+(2+2)+(3+2)+…+(n+2)=(1+2+3+…+n)+2n=n(n+1)2+2n=12n(n+5),故...
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