等差数列前n项和Sn,且2Sn=2k+n^2+n,求k

显示全部楼层 · 2013-5-19 23:42:25

2sn=2k+n^2+n2s1=2a1=2+2k,a1=1+k 2sn-2sn-1=2an=2nan=na1=1=1+kk=0 不一定对，思路大概是这样，答案可能算错数了，没啥自信啊...

千问 · 2013-5-19 23:42:25

解：设：等差数列的首项为a1，公差为dSn=[2a1+(n-1)d]n/22Sn=[2a1+(n-1)d]n依题意和已知，有：[2a1+(n-1)d]n=2k+n&#178+n2(a1)n+dn&#178-dn=2k+n&#178+n2k=2(a1)n+dn&#178-dn-n&#178-n2k=[2(a1)-d-1]n+(d...

千问 · 2013-5-19 23:42:25

设首项为a1,公差为dSn=a1n+n(n-1)d/22a1n+n(n-1)d=n^2d+n(2a1-d)=2k+n^2+nd=1, 2a1-d=1,k=0验算：a1=1an=nSn=(n+1)n/22Sn=n(n+1) =》 k=0...

千问 · 2013-5-19 23:42:25

2Sn=2k+n^2+n
①k为常数2S(n+1)=2k+(n+1)2+(n+1) ②②-①:2a(n+1)=2n+2∴a(n+1)=n+1∴an=n(n≥2)∵{an}为等差数列∴n=1时，a1=1,an=n仍成立那么Sn=(1+n)*n/22Sn=n2+n与①对比∴k=0...

千问 · 2013-5-19 23:42:25

设首项为a1,公差为dSn=a1n+n(n-1)d/22a1n+n(n-1)d=n^2d+n(2a1-d)=2k+n^2+nd=1, 2a1-d=1,k=0...