Question

设各项均为正数的数列{an}的前n项和为Sn，且满足Sn=[（an+1）\2 ]^2求a1a2a3. 并证明数列{an}为等差数列

Answer 1

a1 = S1 = [(a1 + 1) \ 2 ]^2，a1为正，a1 = 1;
a1 + a2 = S2 = [(a2 + 1) \ 2 ]^2，a2为正，a2 = 3;
a1 + a2 + a3 = S3 = [(a3 + 1) \ 2 ]^2，a3为正，a2 = 5;
Sn=[(an + 1) \ 2 ]^2
S(n+1)=[(a(n+1) + 1) \ 2 ]^2
两式相减
a(n+1) = S(n+1) - S(n) = [(a(n+1) + 1) \ 2 ]^2 - [(an + 1) \ 2 ]^2
移项，即
[(a(n+1) - 1) \ 2 ]^2 = [(an + 1) \ 2 ]^2
因为an为正项数列，故a(n+1) - 1 = an + 1，即a(n+1) = an + 2
等差数列

追问

a1 = S1 = [(a1 + 1) \ 2 ]^2，a1为正，a1 = 1;
a1 + a2 = S2 = [(a2 + 1) \ 2 ]^2，a2为正，a2 = 3;
a1 + a2 + a3 = S3 = [(a3 + 1) \ 2 ]^2，a3为正，a2 = 5;
我怎么不理解，能解释一下吗，谢谢您。

追答

对于一个数列，Sn表示前n项和，因此Sn=a1 + a2 + …… + an
因此当n = 1时，S1 = a1；当n = 2时，S2 = a1 +a2；当n = 3时，S3 = a1 + a2 + a3
对于你说的第一行，那是一元二次方程（a1未知），解出来a1 = 1

第二行第三行也是相应的一元二次方程（代入前面求的a1,a2），它们的解一正一负，由于an为正项数列，因此取正数的那个解