已知数列an满足a1=1\2 an+1=an+1\4n平方-1 则an
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已知数列an满足a1=1\2 an+1=an+1\4n平方-1 则an
![已知数列an满足a1=1\2 an+1=an+1\4n平方-1 则an](/uploads/image/z/5283169-25-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%5C2+an%2B1%3Dan%2B1%5C4n%E5%B9%B3%E6%96%B9-1+%E5%88%99an)
a1=1/2
a(n+1)=an+1/(4n²-1)=an+(1/2)[1/(2n-1)-1/(2n+1)]
2a(n+1)=2an+1/(2n-1)-1/(2n+1)
2a(n+1)+1/(2(n+1)-1)=2an+1/(2n-1)
所以 {2an+1/(2n-1)} 为等比数列 首项2a1+1=2 公比q=1 (估计就是每一项都一样的那种咯)
2an+1/(2n-1)=2a1+1=2
an=1-1/2(2n-1)
a(n+1)=an+1/(4n²-1)=an+(1/2)[1/(2n-1)-1/(2n+1)]
2a(n+1)=2an+1/(2n-1)-1/(2n+1)
2a(n+1)+1/(2(n+1)-1)=2an+1/(2n-1)
所以 {2an+1/(2n-1)} 为等比数列 首项2a1+1=2 公比q=1 (估计就是每一项都一样的那种咯)
2an+1/(2n-1)=2a1+1=2
an=1-1/2(2n-1)
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