数列an满足a1=1/3,an+1=an^2+an,记sn=1/(a1+1)+1/(a2+1)+.+1/(1+an),求s10的整_数学解答

数列an满足a1=1/3,an+1=an^2+an,记sn=1/(a1+1)+1/(a2+1)+.+1/(1+an),求s10的整数部分?

人气：192 ℃　时间：2020-05-14 11:48:34

解答

∵a(n+1)=an(an+1)∴1/[a(n+1)]=1/an×1/（an+1)=1/an-1/（an+1)∴1/（an+1)=1/an-1/[a(n+1)]∴s10=1/a1-1/a2+1/a2-1/a3+……+1/a10-1/a11=1/a1-1/a11∵a1=1/3 a2=4/9 ……a4＞1 a(n+1)＞an∴a11＞1 ∴0＜1/a11＜1∵...