Question

设Sn是正项数列{an}的前n项和,且Sn=1/4an^2+1/2an-3/4,求数列{an}通项公式

Answer 1

Sn=An^2/4 + An/2 - 3/4
S(n-1) = [A(n-1)]^2 /4 + [A(n-1)]/2 - 3/4
两式相减得:
An = An^2/4 + An/2 - [A(n-1)]^2 /4 - [A(n-1)]/2
(1/4)[An + A(n-1)][An - A(n-1)] - (1/2)[An + A(n-1)] = 0
因为是正项数列,所以An + A(n-1)不等于0,则
(1/4)[An - A(n-1)]= 1/2
An - A(n-1) = 2
可见这是一个公差为2的等差数列

求出A1：
由Sn=1/4an^2+1/2an-3/4得：
A1 = （1/4）A1^2 + (1/2)A1 - 3/4
(A1 - 3)(A1 + 1) = 0
因A1>0,所以
A1 = 3

An = A1 + (n-1)d = 3 + 2(n - 1) = 2n + 1
数列{an}通项公式：an = 2n + 1

Answer 2

a1=S1=a1^2/4+a1/2-3/4
a1=3
3+a2=a2/4+a2/2-3/4
a2=5

an=2n+1
Sn=(2n+4)*n/2=n(n+2)
an^2/4+an/2=n^2+n+1/4+n+1/2=n^2+2n+3/4