数列an满足a1=1/2,a(n+1)=an^2+an,用[xn]表示不超过x的最大整数见图片,请详尽回答
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![数列an满足a1=1/2,a(n+1)=an^2+an,用[xn]表示不超过x的最大整数见图片,请详尽回答](/uploads/image/z/10909904-32-4.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2F2%2Ca%28n%2B1%29%3Dan%5E2%2Ban%2C%E7%94%A8%5Bxn%5D%E8%A1%A8%E7%A4%BA%E4%B8%8D%E8%B6%85%E8%BF%87x%E7%9A%84%E6%9C%80%E5%A4%A7%E6%95%B4%E6%95%B0%E8%A7%81%E5%9B%BE%E7%89%87%2C%E8%AF%B7%E8%AF%A6%E5%B0%BD%E5%9B%9E%E7%AD%94)
数列an满足a1=1/2,a(n+1)=an^2+an,用[xn]表示不超过x的最大整数见图片,请详尽回答
数列an满足a1=1/2,a(n+1)=an^2+an,用[xn]表示不超过x的最大整数
见图片,请详尽回答
数列an满足a1=1/2,a(n+1)=an^2+an,用[xn]表示不超过x的最大整数见图片,请详尽回答
1/a(n+1)=1/(an^2+an)=1/an-1/(an+1)
1/(an+1)= 1/an-1/a(n+1)
1/(a1+1)+1/(a2+1)+...+1/(a2011+1)=(1/a1-1/a2)+(1/a2-1/a3)+...+(1/a2011-1/a2012)
=1/a1 - 1/a2012=2-1/a2012
因为a(n+1)=an^2 +an
所以a(n+1) -an=an^2 >0
所以{an}是递增数列,
而a2=3/4 a3=21/16
当n>3时,an>a3=21/16
所以0
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