等差数列的题目

\u7b49\u5dee\u6570\u5217\u7684\u9898\u76ee

\u7531\u9898\u610f\u5f97\uff1a
2012\u5e74\u5bb6\u5ead\u7ecf\u8425\u6536\u5165\u4e3a1800+1000\u00d73=4800\u5143
\u5176\u4ed6\u6536\u5165\u4e3a1350+160\u00d73=1830\u5143
\u6240\u4ee52012\u5e74\u8be5\u5730\u533a\u519c\u6c11\u4eba\u5747\u6536\u5165\u4e3a4800+1830=6630\u5143
\u6709\u7591\u95ee\uff0c\u53ef\u8ffd\u95ee\uff1b\u6709\u5e2e\u52a9\uff0c\u8bf7\u91c7\u7eb3\u3002\u795d\u5b66\u4e60\u8fdb\u6b65\u3002

a2\u3001a5\u3001a8\u3001\u2026\u2026\u3001a26\u4e5f\u662f\u4e00\u4e2a\u7b49\u5dee\u6570\u5217\uff0c\u516c\u5deed'=3d=6
a2=a1+d=2+d\uff0ca26=a1+25d=a1+50
\u90a3\u4e48a2+a5+a8+\u2026\u2026+a26=9(a2+a26)/2=9(a1+2+a1+50)/2=9(a1+26)=90
\u6240\u4ee5a1+26=-10\uff0c\u4e8e\u662fa1=-16\uff0c\u90a3\u4e48an=a1+(n-1)d=-16+2(n-1)=2n-18
\u4ee4an=2n-18\u22650\uff0c\u90a3\u4e48n=9\uff0c\u5373\u4ece\u7b2c9\u9879\u5f00\u59cb\uff0can\u5373\u4e3a\u975e\u8d1f\u6570
\u6240\u4ee5Sn\u6709\u6700\u5c0f\u503c\uff0c\u6700\u5c0f\u503c\u4e3aS8\u6216S9(\u8fd9\u4e2a\u662f\u56e0\u4e3aa9=0)
\u800cSn=na1+n(n-1)d/2\uff0c\u6240\u4ee5S8=-16\u00d78+8\u00d77=-72

  解(1)证明  an+1 -1=(an -1)/(an -1)
  1/(an+1 -1)= 1/(an -1)-1
  1/(an+1 -1) - 1/(an -1)=-1
  所以,{1/(an -1)}是以1/(a1 -1)为首项,-1为公差的等差数列
  1/an -1=1/(a1 -1)+(n-1).-1
  an=(n+2)/(n+3)
  (2)bn=1-an
  =1/(n+3)
  sn=b1b2+b2b3+........bnbn+1
  sn=1/4.1/5+1/5.1/6+.....1/n+3.1/n+4
  sn=1/4-1/5+1/5-1/6+....1/n-1/n+4
  sn=1/4-1/n+4
  8sn=2-8/n+4
  8sn=2n/n+4
  8sn-an=2n/n+4-(n+2)/(n+3)
  =n*2-8/(n+3)(n+4)
  所以,当n=1或2时,8sn<an,当n>2时,8sn>an.

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