怎样简便计算?(√1+1/1²+1/2²)+(√1+1/2²+1/3²)+(√1
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/06 10:31:28
怎样简便计算?
(√1+1/1²+1/2²)+(√1+1/2²+1/3²)+(√1+1/3²+1/4²)+...+(√1+1/99²+1/100²)
(√1+1/1²+1/2²)+(√1+1/2²+1/3²)+(√1+1/3²+1/4²)+...+(√1+1/99²+1/100²)
S=√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+√(1+1/3²+1/4²)+...+√(1+1/99²+1/100²)
通项an=√[1+1/n²+1/(n+1)²]
=√{[(n+1)²n²+(n+1)²+n²]/[n²(n+1)²]}
=1/[n(n+1)]* √[n²(n+1)²+2n²+2n+1]
=1/[n(n+1)]*√[n²(n+1)²+2n(n+1)+1]
=1/[n(n+1)]*√[n(n+1)+1]²
=[n(n+1)+1]/[n(n+1)]
=1+1/[n(n+1)]
=1+1/n-1/(n+1)
∴S=a1+a2+a3+.+a99
=(1+1-1/2)+(1+1/2-1/3)+(1+1/3-1/4)+.+(1+1/99-1/100)
=99+(1-1/2+1/2-1/3+1/4-1/4+.+1/99-1/100)
=99+1-1/100
=99又99/100
通项an=√[1+1/n²+1/(n+1)²]
=√{[(n+1)²n²+(n+1)²+n²]/[n²(n+1)²]}
=1/[n(n+1)]* √[n²(n+1)²+2n²+2n+1]
=1/[n(n+1)]*√[n²(n+1)²+2n(n+1)+1]
=1/[n(n+1)]*√[n(n+1)+1]²
=[n(n+1)+1]/[n(n+1)]
=1+1/[n(n+1)]
=1+1/n-1/(n+1)
∴S=a1+a2+a3+.+a99
=(1+1-1/2)+(1+1/2-1/3)+(1+1/3-1/4)+.+(1+1/99-1/100)
=99+(1-1/2+1/2-1/3+1/4-1/4+.+1/99-1/100)
=99+1-1/100
=99又99/100
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