计算1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+50)

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计算1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+50)
计算1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+50)

计算1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+50)
1+1/[(1+2)*2/2]+1/[(1+3)*3/2]+.+1/[(1+50)*50/2]
=1+2/(2*3)+2/(3*4)+2/(4*5)+.+2/(50*51)
=1+2(1/2*3+1/3*4+1/4*5+.+1/50*51)
=1+2(1/2-1/51)
=1+49/51
=100/51

1+1/[(1+2)*2/2]+1/[(1+3)*3/2]+1/[(1+4)*4/2]+......+1/[(1+50)*50/2]
=1+2/1*2+2/2*3+2/3*4+2/4*5+......2/50*51
=1+2*(1-1/2+1/2-1/3+....+1/50-1/51)
=1+2*(1-1/51)
=3-2/5
=13/5

先把分母一个一个的算出来,再找分母有倍数关系的分数相加,就好算了

n(n+1)的倒数裂项为n的倒数和n+1的倒数差依次消项我打字不好你自己算吧