一道等差数列的题
热心网友
1.设bn为等差级数。2b(n)=b(n-1)+b(n+1)an=[(n+1)/2]bn-[(n-1)/2]b(n-1)==a(n+1)+a(n-1)=[(n+2)/2]b(n+1)-[(n)/2]b(n)+[(n)/2]b(n-1)-[(n-2)/2]b(n-2)=[(n+2)/2][2b(n)-b(n-1)]-[(n)/2]b(n)+[(n)/2]b(n-1)--[(n-2)/2][2b(n-1)-bn]=2[(n+1)/2]bn-2[(n-1)/2]b(n-1)==2an ，所以an为等差级数。2.设an=a+k（n-1）==bn=a+k[2*1+3*2+。。+n（n-1）]/[n（n+1）/2]==a+k[（n+1）n（n-1）/3]/[n（n+1）/2]=a+[2k/3]（n-1）==所以bn为等差级数。
热心网友
设:{bn}为等差数列,公差为d,b(n+1)-bn=[a1+2a2+3a3+。。。。+(n+1)a(n+1)]/(1+2+3+。。。+n+n+1)-[a1+2a2+3a3+。。。。+nan]/(1+2+3+。。。+n)=[a1+2a2+3a3+。。。。+(n+1)a(n+1)]/[(n+2)(n+1)/2]-[a1+2a2+3a3+。。。。+nan]/[n(n+1)/2]=2n[a1+2a2+3a3+。。。。+(n+1)a(n+1)]-2(n+2)[a1+2a2+3a3+。。。。+nan]/[n(n+1)(n+2)]=[2n(n+1)a(n+1)-4(a1+2a2+3a3+。。。nan)]/[n(n+1)(n+2)]=2a(n+1)/(n+2)-2bn/(n+2)=d;a(n+1)-bn=d(n+2)/2;a(n+1)=d(n+2)/2+bn;an=d(n+1)/2+b(n-1);a(n+1)-an=d/2+bn-b(n-1)=3d/2;则:{an}为等差数列设:{an}为等差数列,公差为k,b1=a1;b2=[a1+2(a1+k)]/3=[3a1+2k]/3=b1+2k/3;b3=[a1+2(a1+k)+3(a1+2k)]/(1+2+3)=b1+4k/3;用数学归纳法可证:{bn}为等差数列,公差为2k/3。