若等差数列{an)的前n项和为Sn,公差为d, 则Sn=na1+1/2n(n-1)d =d/2n2+(a1-d/2)n. 令a=d/2,b=a1-d/2,于是Sn=an2+bn(n=1,2,…). 例1 等差数列的S10=20,S20=60,则S30的值是____. (第四届93年“希望杯”高二1试) 解设前n项和Sn=an2+bn,由题设有(?)20=100a+10b,60=400a+20b.解得(?)a=1/10,b=1. 所以S30=900×1/10+30=120. 例2 已知数列{an)为等差数列,若 If the sum of the first n terms of the equal difference sequence {an) is Sn and the tolerance is d, then Sn=na1+1/2n(n-1)d=d/2n2+(a1-d/2)n. Let a=d /2,b=a1-d/2, then Sn=an2+bn(n=1,2,...). Example 1 In the arithmetic progression S10=20, S20=60, then the value of S30 is ____. ( The fourth session of the “Hope Cup” in the second year of the “I hope cup” 1) Solve the first n items and Sn = an2 + bn, by the title with (?) 20 = 100a + 10b, 60 = 400a + 20b. Solution (?) a =1/10, b=1. So S30=900×1/10+30=120. Example 2 Known sequence {an) is equal arithmetic sequence if