数列{an}的前n项和Sn = n2+2n+5,则a6+a7+a8 =

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数列{an}的前n项和Sn=n2+2n+5,则a6+a7+a8=数列{an}的前n项和Sn=n2+2n+5,则a6+a7+a8=数列{an}的前n项和Sn=n2+2n+5,则a6+a7+a8=a6+a

数列{an}的前n项和Sn = n2+2n+5,则a6+a7+a8 =
数列{an}的前n项和Sn = n2+2n+5,则a6+a7+a8 =

数列{an}的前n项和Sn = n2+2n+5,则a6+a7+a8 =
a6+a7+a8
=S8-S5
=8²+2×8+5-(5²+2×5+5)
=45