√(1+(1/1^2)+(1/1^2))+√(1+(1/2^2)+(1/3^2))+√(1+(1/3^2)+(1/4^2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 10:28:21
√(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)
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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