计算下面各题.(能简算的要简算)谢谢,急.(1)9.81×0.1+0.5×98.1+0.049×981(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1

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计算下面各题.(能简算的要简算)谢谢,急.(1)9.81×0.1+0.5×98.1+0.049×981(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1
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计算下面各题.(能简算的要简算)谢谢,急.(1)9.81×0.1+0.5×98.1+0.049×981(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1
计算下面各题.(能简算的要简算)谢谢,急.
(1)9.81×0.1+0.5×98.1+0.049×981
(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1

计算下面各题.(能简算的要简算)谢谢,急.(1)9.81×0.1+0.5×98.1+0.049×981(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1
.(1)9.81×0.1+0.5×98.1+0.049×981
=9.81x0.1+5x9.81+4.9x9.81
=9.81x(0.1+5+4.9)
=9.81x10
=98.1
(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1
=1-1/2+1/2-1/3+.+1/199-1/200
=1-1/200
=199/200

9.81×0.1+0.5×98.1+0.049×981=98.1*0.01+98.1*0.5+98.1*0.49=1*98.1=98.1
原式=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5……+1/199-1/200=1-1/200=199/200(用裂项法)

(1)9.81×0.1+0.5×98.1+0.049×981=9.81×0.1+5×9.81+4.9×9.81=9.81×(0.1+5+4.9)=9.81×10=98.1
(2)1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1=(1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/199-1/200)=1-1/200=199/200

9.81×0.1+0.5×98.1+0.049×981
=9.81×0.1+9.81×5+9.81×4.9
=9.81×(0.1+4.9+5)
=9.81×10
=98.1

1/1-1/2+1/2-1/3+1/3-1/4+1/5-1/5+.....+1/199-1/200
=1-1/200
=199/200

(1)9.81×0.1+0.5×98.1+0.049×981
=9.81×0.1+5×9.81+4.9×9.81
=9.81*(0.1+5+4.9)
=98.1
(2)1*(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(199*200)
=1-1/2-+1/2-1/3+1/3-1/4+……1/199-1/200
=1-1/200
=199/200

(1)9.81×0.1+0.5×98.1+0.049×981
=9.81×0.1+5×9.81+4.9×9.81
=9.81×(0.1+5+4.9)
=9.81×10
=98.1
(2)
1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1
=1/1-1/2+1/2-1/3+1/3-1/4...

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(1)9.81×0.1+0.5×98.1+0.049×981
=9.81×0.1+5×9.81+4.9×9.81
=9.81×(0.1+5+4.9)
=9.81×10
=98.1
(2)
1×2分之1+2×3分之1+3×4分之1+4×5分之1+……+199×200分之1
=1/1-1/2+1/2-1/3+1/3-1/4+1/4……+1/199-1/200
=1-1/200
=199/200


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1、原式=9.81×0.1+9.81×5+9.81×4.9=9.81×10=98.1;
2、路上原来是初等题目啊,如果原式=1×0.5+2×(3分之1)+3×(4分之1)+4×(5分之1)+……+199×(200分之1),则设为数列,则a1=1/2,a2=2/3...an=n/(n+1)=1-1/(n+1),故Sn=n-1/[2(n-1)!]=(2n!-1)/2/(n-1)!,n=200时...

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1、原式=9.81×0.1+9.81×5+9.81×4.9=9.81×10=98.1;
2、路上原来是初等题目啊,如果原式=1×0.5+2×(3分之1)+3×(4分之1)+4×(5分之1)+……+199×(200分之1),则设为数列,则a1=1/2,a2=2/3...an=n/(n+1)=1-1/(n+1),故Sn=n-1/[2(n-1)!]=(2n!-1)/2/(n-1)!,n=200时,Sn=200-1/398!

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