S=1-1/100=99/100

bạn tách ra, 1/1.2=1-1/2 cứ như thế, rồi trừ đi còn 1-1/100=99/100

S = 99/100

Giải:

Ta có: \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}...+\dfrac{1}{99.100}\)

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

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

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

\(\Leftrightarrow S=\dfrac{99}{100}\)

Vậy ...

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

S=1-1/100=99/100

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101

=> 3S = 99.100.101

=> S = \(\frac{99.100.101}{3}=333300\)

ta xét

\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)

\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)

\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)

Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)

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

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

=1-1/100

=99/100

=1−1/2+1/2−1/3+1/3−1/4+...+1/99−1/100
=1 − 1/100 = 99/100

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

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

= 1 - 1/100

= 99/100

1. ta có :

\(3^2+4^2=5^{x-1}\)

  \(25=5^{x-1}\)

 \(5^2=5^{x-1}\)

=> x = 3

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101

=> 3S = 99.100.101

=> S = 99.100.101/3

=> S = 333300 

v

Answer:

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

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

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

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

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

5050 đấy bạn mình cũng không chắc lắm

 S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 + 2.3(4 - 1) + 3.4(5 - 2) + ......... + 99.100(101 - 98)

=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ........ + 99.100.101 - 98.99.100

=> 3S = (1.2.3 + 2.3.4 + 3.4.5 + ..... + 98.99.100 + 99.100.101) - (1.2.3 + 2.3.4 + .......... + 98.99.100)

=> 3S = 99.100.101

=> S = \(\frac{99.100.101}{3}=333300\)

Đặt S = 1 x 2 + 2 x 3 + 3 x 4 +... + 99 x 100

3 S = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 98 x 99 x 3 + 99 x 100 x 3

3 S = 1 x 2 x 3 + 2 x 3 ( 4 - 1 ) + 3 x 4 ( 5 - 2 ) + ... + 98 x 99 ( 100 - 97 ) + 99 x 100 ( 101 - 98 )

3 S = 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 + ... - 97 x 98 x 99 + 99 x 100 x 101 - 98 x 99 x 100

3 S = 99 x 100 x 101  3S = 3 x 33 x100 x 101

S = 33 x 100 x 101 = 333 300