√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)……初二数学√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√(1+1/2009^2+1/2010^2)要步骤,
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 01:28:44
![√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)……初二数学√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√(1+1/2009^2+1/2010^2)要步骤,](/uploads/image/z/4101929-17-9.jpg?t=%E2%88%9A%281%2B1%2F12%2B1%2F%5E2%29%2B%E2%88%9A%281%2B1%2F2%5E2%2B1%2F3%5E2%29%E2%80%A6%E2%80%A6%E5%88%9D%E4%BA%8C%E6%95%B0%E5%AD%A6%E2%88%9A%281%2B1%2F12%2B1%2F%5E2%29%2B%E2%88%9A%281%2B1%2F2%5E2%2B1%2F3%5E2%29%2B%E2%88%9A%281%2B1%2F3%5E2%2B1%2F4%5E2%29%E2%80%A6%E2%80%A6%E2%88%9A%EF%BC%881%2B1%2F2009%5E2%2B1%2F2010%5E2%EF%BC%89%E8%A6%81%E6%AD%A5%E9%AA%A4%2C)
√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)……初二数学√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√(1+1/2009^2+1/2010^2)要步骤,
√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)……初二数学
√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√(1+1/2009^2+1/2010^2)要步骤,
√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)……初二数学√(1+1/12+1/^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√(1+1/2009^2+1/2010^2)要步骤,
√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√(1+1/2009^2+1/2010^2)
√(1+1/1^2+1/2^2)= √(1+1+1/4)= √9/4=3/2=1+1/2
√(1+1/2^2+1/3^2)= √(1+1/4+1/9)= √49/36=7/6=1+1/(2*3)
√(1+1/3^2+1/4^2)=1+1/3*4
……
√(1+1/2009^2+1/2010^2)=1+1/2009*2010
设√(1+1/a^2+1/(a+1)^2)=1+1/a(a+1)
右边平方得:[a(a+1)+1]^2/[a(a+1)]^2=[a(a+1)^2+2a(a+1)+1}/[a(a+1)]^2=1+(2a^2+2a+1)/ [a(a+1)]^2=1+[(a^2+2a+1)+a^2]/[a(a+1)]^2=1+[(a+1)^2+a^2] /[a(a+1)]^2=1+1/a^2+1/(a+1)^2与根号下相同所以等式成立.
1/a-1/(a+1)=(a+1-a)/a(a+1)=1/a(a+1)
所以1/a(a+1)=1/a-1/(a+1)成立
原式=1+1/1*2+1+1/2*3+……+1+1/2009*2010=2009+1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010=2009+1-1/2010=2010-1/2010=2009/2010