\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{1}-\frac{1}{100}\)

\(\frac{100-1}{100}\)

\(\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)

 

Ta có :

\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+..............+\dfrac{1}{99.100}\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)

\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}\)

=\(\frac{99}{100}\)

ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\) 

          \(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

           \(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)

            ......

          \(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

=> \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=>A=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}+\frac{1}{100\cdot101}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

\(=1+1-\frac{1}{101}=2-\frac{1}{101}=1\frac{100}{101}=\frac{201}{101}\)

=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

1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

1 \(\times\) 2 \(\times\) 3 = 1  \(\times\) 2 \(\times\) 3

2 \(\times\) 3 \(\times\) 3 = 2 \(\times\) 3 \(\times\) ( 4 -1) = 2  \(\times\) 3 \(\times\) 4 - 1 \(\times\) 2 \(\times\) 3

3 \(\times\) 4 \(\times\) 3 = 3 \(\times\) 4 \(\times\) ( 5 -2) = 3 \(\times\) 4 \(\times\) 5 - 2 \(\times\) 3 \(\times\) 4

4 \(\times\) 5 \(\times\) 3 = 4 \(\times\) 5 \(\times\) ( 6- 3) = 4 \(\times\) 5 \(\times\) 6 - 3 \(\times\) 4 \(\times\) 5 

..................................................................................

99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)(101-98) =99\(\times\)100\(\times\)101 - 98\(\times\)99\(\times\)100

Cộng vế với vế ta được:

1\(\times\)2\(\times\)3 + 2\(\times\)3\(\times\)3 + 3\(\times\)4\(\times\)3+ ...+99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)101

(1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4 +...+99\(\times\)100)\(\times\)3 = 99\(\times\)100\(\times\)101

1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = (99 \(\times\)100 \(\times\)101):3

1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = 333 300

cảm ơn cô

3M = 1.2.3 + 2.3.(4-1)  +..+  99.100.(101-98)

3M = 1.2.3 + 2.3.4 - 1.2.3 + .... + 99.100.101 - 98.99.100

3M = 99.100.101 = 999900

M  = 333300

333300

tick nha,nha!!!!!!!!!!!!!!!!!!!!!

A = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100

3A = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + 99 x 100 x (101 - 98)

3A = 1 x 2 x 3 - 0 x 1 x 2 + 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

3A = (1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ... + 99 x 100 x 101) - (0 x 1 x 2 + 1 x 2 x 3 + 2 x 3 x 4 + ... + 98 x 99 x 100)

3A = 99 x 100 x 101

A = 33 x 100 x 101

A = 333300

A = 1x2+2x3+3x4+4x5+....+99x100

3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+...+99x100x(101-98)

3A= 1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+...+99x100x101-98x99x100

3A= 99x100x101

A=999900 : 3 = 333300