√(1+(1/1^2)+(1/1^2))+√(1+(1/2^2)+(1/3^2))+√(1+(1/3^2)+(1/4^2))+……+√(1+(1/2007^2)+(1/2008^2))
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 03:28:25
![√(1+(1/1^2)+(1/1^2))+√(1+(1/2^2)+(1/3^2))+√(1+(1/3^2)+(1/4^2))+……+√(1+(1/2007^2)+(1/2008^2))](/uploads/image/z/1690866-18-6.jpg?t=%E2%88%9A%EF%BC%881%2B%EF%BC%881%2F1%5E2%29%2B%281%2F1%5E2%29%29%2B%E2%88%9A%281%2B%281%2F2%5E2%29%2B%281%2F3%5E2%29%29%2B%E2%88%9A%EF%BC%881%2B%EF%BC%881%2F3%5E2%29%2B%281%2F4%5E2%EF%BC%89%EF%BC%89%2B%E2%80%A6%E2%80%A6%2B%E2%88%9A%281%2B%281%2F2007%5E2%29%2B%281%2F2008%5E2%29%29)
√(1+(1/1^2)+(1/1^2))+√(1+(1/2^2)+(1/3^2))+√(1+(1/3^2)+(1/4^2))+……+√(1+(1/2007^2)+(1/2008^2))
√(1+(1/1^2)+(1/1^2))+√(1+(1/2^2)+(1/3^2))+√(1+(1/3^2)+(1/4^2))+……+√(1+(1/2007^2)+(1/2008^2))
√(1+(1/1^2)+(1/1^2))+√(1+(1/2^2)+(1/3^2))+√(1+(1/3^2)+(1/4^2))+……+√(1+(1/2007^2)+(1/2008^2))
√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+√(1+1/3²+1/4²)+……+√(1+1/2007²+1/2008²)
先推导公式:√[1+1/n²+1/(n+1) ² = √{[n² (n+1)²+n²+(n+1)²]/[n² (n+1)²]}= √{ (n²+n+1) ²/ [n² (n+1)²]}= (n²+n+1)/ [n (n+1)]=1+1/n-1/(n+1)
故:√(1+1/1²+1/2²)=1+1/1-1/2
√(1+1/2²+1/3²)=1+1/2-1/3
√(1+1/3²+1/4²)=1+1/3-1/4
……
√(1+1/2007²+1/2008²)=1+1/2007-1/2008
故:√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+√(1+1/3²+1/4²)+……+√(1+1/2007²+1/2008²)
=1+1/1-1/2+1+1/2-1/3+1+1/3-1/4+….+ 1+1/2007-1/2008
=1×2007+1/1-1/2008
=2007又2007/2008 (即:2007+2007/2008)
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