1/1x2 + 1/2x3 + 1/3x4 + ... + 1/99x100 + 1/100x101 = ...
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\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}+\frac{1}{100\cdot101}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
\(=1+1-\frac{1}{101}=2-\frac{1}{101}=1\frac{100}{101}=\frac{201}{101}\)
=1+1/1-1/2+1/2-1/3+1/3-1/+1/4-1/5+...+1/99-1/100+1/100-1/101
=1+1-1/101
=201/101
Đây là câu trả lời của mình :
Đặt A = 1 x 2 + 2 x 3 + 3 x 4 + ........ + 99 x 100 + 100 x 101
3A = 1 x 2 x3 + 2 x 3 x 3 + 3 x 4 x 3 + ........... + 99 x 100 x 3 + 100 x 101 x 3
3A = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ........... + 99 x 100 x ( 101 - 98 ) + 100 x 101 x ( 102 - 99 )
3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .......... + 99 x 100 x 101 - 98 x 99 x 100 + 100 x 101 x 102 - 99 x 100 x 101
3A = 100 x 101 x 102
3A = 1030200
A = 1030200 : 3
A = 343400
Vậy : 1 x 2 + 2 x 3 + 3 x 4 + ......... + 99 x 100 + 100 x 101 = 343400
A=1x2+2x3+3x4+4x5+......+99x100+100x101
3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+99x100x(101-98)+100x101x(102-99)
3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+...+99x100x101-98x99x100+100x101x102-99x100x101
3A=(1x2x3+2x3x4+3x4x5+4x5x6+...+99x100x101+100x101x102)-(0x1x2+1x2x3+2x3x4+3x4x5+...+98x99x100+99x100x101)
3A=100x101x102
A=100x101x102:3
A=343400
A = 1x2 + 2x3 + 3x4 + 4x5 + ... + 99x100 + 100x101
3A = 1x2x(3-0) + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) + 100x101x(102-99)
3A = 1x2x3 - 0x1x2 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100 + 100x101x102 - 99x100x101
3A = 100x101x102 - 0x1x2
3A = 100x101x102
A = 100x101x34
A = 343400
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy.....
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(A=1+0-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}< 1\)
\(\Rightarrow A< 1\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)