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(1) 1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)
(2) abc=1 ,计算 1/(1+a+ab)+1/(1+b+bc)+1/(1+c+ca)
人气:339 ℃ 时间:2020-06-27 05:13:24
解答
(1) 1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)
=2/1*2+2/2*3+2/3*4+.+2/100*101
=2(1/1*2+1/2*3+1/3*4+.1/100*101)
=2(1/1-1/2+1/2-1/3+1/3-1/4+.+1/100-1/101)
=2*(1-1/101)
=200/101
(2) abc=1 ,计算
1/(1+a+ab)+1/(1+b+bc)+1/(1+c+ca)
=abc/(abc+a+ab)+abc/(1+b+bc)+1/(1+c+ca)
=bc/(bc+1+b)+abc/(1+b+bc)+1/(1+c+ca)
=(bc+abc)/(bc+1+b)+1/(1+c+ac)
=b(c+ac)/(bc+abc+b)+1/(1+c+ac)
=(c+ac)/(c+ac+1)+1/(1+c+ac)
=(c+ac+1)/(c+ac+1)
=1
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