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bn+1=2bn²-bn+1/2求通项!我要的是方法!公式!已知f(x)=x²-1/2x+1/4若{bn}满足b1=b,bn+1=2f(bn)其中“bn+1”为b的第n+1项(1)当b=1/2时,数列{bn}是否为等差数列?若是请求出通项(2)当1/2<
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bn+1=2bn²-bn+1/2求通项!我要的是方法!公式!
已知f(x)=x²-1/2x+1/4若{bn}满足b1=b,bn+1=2f(bn) 其中“bn+1”为b的第n+1项
(1)当b=1/2时,数列{bn}是否为等差数列?若是请求出通项
(2)当1/2<b<1时求证1/b1+1/b2+1/b3+.+1/bn
已知f(x)=x²-1/2x+1/4若{bn}满足b1=b,bn+1=2f(bn) 其中“bn+1”为b的第n+1项
(1)当b=1/2时,数列{bn}是否为等差数列?若是请求出通项
(2)当1/2<b<1时求证1/b1+1/b2+1/b3+.+1/bn
▼优质解答
答案和解析
(1)当bn = b = 1/2时,b(n+1) = b,所以对任意n,bn = b = 1/2
(2)b(n+1) = 2bn^2 - bn + 1/2
所以b(n+1) - 1/2 = 2bn^2 - bn
1/[ b(n+1) - 1/2 ] = 1/[ 2bn (bn - 1/2) ]
1/[ b(n+1) - 1/2 ] = 1/[ bn - 1/2 ] - 1/bn
所以 1/bn = 1/[ bn - 1/2 ] - 1/[ b(n+1) - 1/2 ]
1/b1+1/b2+1/b3+.+1/bn = 1/[ b1 - 1/2] - 1/[ b(n+1) - 1/2 ] < 1/[ b -1/2] = 2/(2b-1)
(2)b(n+1) = 2bn^2 - bn + 1/2
所以b(n+1) - 1/2 = 2bn^2 - bn
1/[ b(n+1) - 1/2 ] = 1/[ 2bn (bn - 1/2) ]
1/[ b(n+1) - 1/2 ] = 1/[ bn - 1/2 ] - 1/bn
所以 1/bn = 1/[ bn - 1/2 ] - 1/[ b(n+1) - 1/2 ]
1/b1+1/b2+1/b3+.+1/bn = 1/[ b1 - 1/2] - 1/[ b(n+1) - 1/2 ] < 1/[ b -1/2] = 2/(2b-1)
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