Question

已知等差数列an中a1=1且 s2n-2sn=n的平方 设bn=n+q的an次方 求bn前n项和tn

Answer 1

∵等差数列an ∴S2n＝n(a1＋a2n) Sn＝n(a1＋an)/2

∵S2n－2Sn＝n² ∴n(a1＋a2n)－2×n(a1＋an)/2＝n²

∴a2n－an＝n ∴nd＝n ∴d＝1

∵等差数列an中a1=1 ∴an＝n

∴bn＝n＋q^n

∴Tn＝(1＋2＋…＋n)＋(q＋q²＋＋…q^n)＝n(n＋1)/2＋q(1－q^n)/(1－q)