一、前n项和S_n中的项数例1已知数列的前n项和S_n=2n~2+3n+3,求a_n=分析此题如果我们仅仅是利用a_n=S_n-S_(n-1)求解那么就有可能求错,因为首项是写不进通项公式的,换句话说该数列是从第二项开始的等差数列。解:由S_n=2n~2+3n+3S_(n-1)=2(n-1)~2+3(n-1)+3可得a_n=4n+1(n≥2)又因为当n=1时,S_1=a_1=8 First, the first n items and the number of terms in S_n Example 1 Known series of first n terms and S_n = 2n ~ 2 + 3n + 3, find a_n = Analysis of the problem if we only use a_n = S_n-S_ (n- 1) Solving then it is possible to make mistakes, because the first item is not written into the formula, in other words the series is from the second start of the arithmetic sequence. Solution: a_n = 4n + 1 (n≥2) is obtained from S_n = 2n ~ 2 + 3n + 3S_ (n-1) = 2 (n-1) ~ 2 + 3 (n-1) +3 When n = 1, S_1 = a_1 = 8