Question

已知数列an为等差数列，Sn为其前n项和，且a2=3.4S2=S4.求数列an的通项公式

Answer 1

4S2=S4 4(a1+a2)=a1+a2+a3+a4 3(a1+a2)=a3+a4 3(a2-d+a2)=a2+d+a2+2d 2a2=3d 又a2=3 所以d=2 a1=1 an=a1+(n-1)d =2n- 1

Answer 2

S2 = 2a1+1dS4 = 4a1 + 6d4S2 = S48a1 + 4d = 4a1 + 6d4a1 = 2d d= 2a1a1 + a2 = S2 = 2a1 + 1d = 2a1 +2a1a1 + 3 = 4a13a1 = 3 a1 =1a1 =1 a2 = 3 所以 d=2an = 1+2(n-1) = 2n-1