设数列{an}满足a1=0且1/(1-an+1)-1/(1-an)=1

显示全部楼层 · 2013-7-9 19:53:19

bn=1/根号n-1/根号(n+1)所以b1=1-1/根号2 b2=1/根号2-1/根号3,b3=1/根号3-1/根号4......sn=b1+b2+b3+.......+bn
=(1-1/根号2)+(1/根号2-1/根号3)+(1/根号3-1/根号4)+.......[1/根号n-1/根号(n+1)] =1-1/根号2+1/根号2-1/根号3+1/根号3-1/根号4+......+1/根号n-1/根号(n+1) =1-1/根号(n+1)<1明白了吗？...

千问 · 2013-7-9 19:53:19

bn=[1-√a(n+1)]/√n＝[1-√[1-1/(n+1)]/√n=[1-√[n/(n+1)]/√n
分子分母乘√(n+1)=[√(n+1)-√n]/[√n√(n+1)]=1/√n-1/√(n+1)Sn=∑bk=1-1/√(n+1)<1 (这里Bk前后项抵消)...

千问 · 2013-7-9 19:53:19

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千问 · 2013-7-9 19:53:19

通项公式怎么算...