√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/16 09:22:05
√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2
√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6
√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12
根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的理由
√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6
√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12
根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的理由
规律:√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))
证明:假设√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))成立
则:(1+1/(n+1)-1/(n+2))^2=1+2/(n+1)-2/(n+2)+1/((n+1)^2)+1/((n+2)^2)-2/((n+1)(n+2))
=1+1/((n+1)^2)+1/((n+1)^2)
即:1+1/((n+1)^2)+1/((n+1)^2)=(1+1/(n+1)-1/(n+2))^2
两边各开根号:得:√1+1/((n+1)^2)+1/((n+1)^2)=1+1/(n+1)-1/(n+2)
又n (n为大于1)的自然数,n=1时等式成立
综上所述:√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))
证明:假设√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))成立
则:(1+1/(n+1)-1/(n+2))^2=1+2/(n+1)-2/(n+2)+1/((n+1)^2)+1/((n+2)^2)-2/((n+1)(n+2))
=1+1/((n+1)^2)+1/((n+1)^2)
即:1+1/((n+1)^2)+1/((n+1)^2)=(1+1/(n+1)-1/(n+2))^2
两边各开根号:得:√1+1/((n+1)^2)+1/((n+1)^2)=1+1/(n+1)-1/(n+2)
又n (n为大于1)的自然数,n=1时等式成立
综上所述:√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))
√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2
1+2+1+2+1+2+1+2+1+2+1+2+1+2 =( )*( ) =()
1*1/2=?
1+1×2=?
1、-|-1/2|=?
1/(1+√2)+1/(√2+√3)+...+1/(√9999+√10000)=99
1 1 1 1------ + ------ + ------ + .+ -------------- =1×2 2×3
1/(1+√2)+1/(√2+√3)+1/(√3+2)+.+1/(3+√10)=
(1\2012-1)(1\2011-1)(1\2010-1)...(1\3-1)(1\2-1)=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
已知S1=1+1/1^2+1/2^2
=2[(1/2)cos2x+(√3/2)sin2x] + 1