√[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
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 06:27:47
![√[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](/uploads/image/z/263093-5-3.jpg?t=%E2%88%9A%5B1%2B1%2F%281%5E2%29%2B1%2F%282%5E2%29%5D+%3D+1%2B1%2F1-1%2F2%3D1%2B1%2F2%E2%88%9A%5B1%2B1%2F%282%5E2%29%2B1%2F%283%5E2%29%5D+%3D+1%2B1%2F2-1%2F3%3D1%2B1%2F6%E2%88%9A%5B1%2B1%2F%283%5E2%29%2B1%2F%284%5E2%29%5D+%3D+1%2B1%2F3-1%2F4%3D1%2B1%2F12%E6%A0%B9%E6%8D%AE%E4%B8%8A%E8%BF%B0%E7%AD%89%E5%BC%8F%E6%8F%AD%E7%A4%BA%E7%9A%84%E8%A7%84%E5%BE%8B%2C%E5%86%99%E5%87%BA%E7%94%A8%E5%AD%97%E6%AF%8Dn+%28n%E4%B8%BA%E5%A4%A7%E4%BA%8E1%29%E7%9A%84%E8%87%AA%E7%84%B6%E6%95%B0%E8%A1%A8%E7%A4%BA%E8%BF%99%E4%B8%80%E8%A7%84%E5%BE%8B%E7%9A%84%E7%AD%89%E5%BC%8F%2C%E5%B9%B6%E8%AF%B4%E6%98%8E%E7%AD%89%E5%BC%8F%E6%88%90%E7%AB%8B%E7%9A%84)
√[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
√[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/(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
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))