高一数学问题
已知在正整数数列{an}中,前n项和sn满足:Sn=1/8(an+2)^2
求证:{an}是等差数列。
已知Sn=(an+2)^2/8
所以,S1=a1=(a1+2)^2/8
===> (a1+2)^2=8a1
===> a1^2+4a1+4=8a1
===> a1^2-4a1+4=0
===> (a1-2)^2=0
===> a1=2
又,S=[a+2]^2/8
所以,Sn-S=[(an+2)^2-(a+2)^2]/8
即,an=[(an+2)^2-(a+2)^2]/8
===> 8an=an^2+4an+4-a^2-4a-4
===> 8an=an^2+4an-a^2-4a
===> ...全部
已知在正整数数列{an}中,前n项和sn满足:Sn=1/8(an+2)^2
求证:{an}是等差数列。
已知Sn=(an+2)^2/8
所以,S1=a1=(a1+2)^2/8
===> (a1+2)^2=8a1
===> a1^2+4a1+4=8a1
===> a1^2-4a1+4=0
===> (a1-2)^2=0
===> a1=2
又,S=[a+2]^2/8
所以,Sn-S=[(an+2)^2-(a+2)^2]/8
即,an=[(an+2)^2-(a+2)^2]/8
===> 8an=an^2+4an+4-a^2-4a-4
===> 8an=an^2+4an-a^2-4a
===> an^2-4an-a^2-4a=0
===> (an^2-a^2)-4(an+a)=0
===> (an+a)*(an-a)-4(an+a)=0
===> (an+a)*(an-a-4)=0
已知数列各项为正整数,即an、a>0
所以,an-a-4=0
所以,an-a=4
即,数列an是以a1=2为首项,公差d=4的等差数列
即,an=a1+(n-1)d=2+(n-1)*4=4n-2
【检验:
根据等差数列求和公式有,Sn=na1+[n(n-1)d]/2
=2n+[n(n-1)*4]/2=2n+2n(n-1)=2n^2
而,题目所给表达式为Sn=(an+2)^2/8=[(4n-2)+2]^2/8
=(4n)^2/8=2n^2
显然是相等的。
】。收起