（1）∵等差数列{an}中a1=1，公差d=1∴Sn＝na1+n(n−1)2d＝n2+n2∴bn＝2n2+n…（4分）（2）∵bn＝2n2+n＝2n(n+1)…（6分）∴b1+b2+b3+…+bn＝2(11×2+12×3+13×4+…+1n(n+1))=2(1−12+12−13+13−14+…+1n−1n+1)…...