已知数列{An}的前N项和Sn，求Sn^(1/2)是等差数列的充要条件

千问 · 2012-1-10 01:30:46

{Sn^(1/2)}是等差数列的充要条件是：原数列{An}也是一个等差数列，并且其公差d=2A1≥0.证明：可设An=A1（n-1）(2A1)，(A1≥0）所以Sn=nA1n(n-1)d/2=A1*n所以Sn^(1/2)=A1^(1/2)*n所以,{Sn^(1/2)}是公差为根号下A1的等差数列.

千问 · 2012-1-10 01:30:46

必要性∵√Sn次为等差数列所以可是设Sn=(b1(n-1)d\')^2a1=b1^2当n1时an=Sn-Sn-1=(b1(n-1)d\'-b1-(n-2)d\')(2b1(2n-3)d\')=d\'(2b1(2n-3)d\')当n2时an-an-1=d\'(2b1(2n-3)d\')-d\'(2b1(2(n-1)-3)d\')=2d\'^2为常数对n2{an}为首项为2b1d\'公差为2d\'^2的等差数列a1=b1^2充分性假设{an}a1=b1^2b1为常数{a(n1)}n0为等差数列首项为2b1d\'公差为2d\'^2Sn=b1^2(2b1d\'2b1d\'(n-1)2d‘^2)*(n-1)/2=b1^2(2b1d\'(n-1)d\'^2)(n-1)=b1^22b1d\'(n-1)(n-1)^2d\'^2=(b1(n-1)d\')^2√Sn-√Sn-1=d\'所以√Sn是等差数列的充要条件是存在常数b1与d\'使得a1=b1^2{a(n1)},n0是以首相为2b1d\'公差为2d\'^2的等差数列~~~~~赞同