已知数列{an}的前n项和为Sn=2n^2,则a3+a4+a5=
问题描述:
已知数列{an}的前n项和为Sn=2n^2,则a3+a4+a5=
答
S5-S2=(a1+a2+a3+a4+a5)-(a1+a2)
=a3+a4+a5
=2*5^2-2*2^2
=50-8=42