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

S=1/1.2+1/2.3+1/3.4+...+1/99.10 

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

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

\(\Rightarrow S=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

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

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

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

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 \(S\left(100\right)=\frac{99.100.101}{3}=333300\)

s=1.2+2.3+3.4+...+99.100

=>3s=1.2.3+2.3.3+3.4.3+...+99.100.3

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

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

=99.100.101

=>s=99.100.101/3=333300

a, S< 2²⁰¹⁸

b, M=1.2+2/1.2+2.3+2/2.3+.....+99.100+2/99.100

M= 2+2+2+2+2+2+.....+2

M=100 vì có 50 số 2

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 

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

  a)    S= 1+ 1/2 + 1/4 +1/8+ …+1/1024

      ½ S=1/2x1+1/2x1/2+1/2x1/4+1/2x1/8+… + 1/1024

            =1/2+1/8+1/16+…+1/1024+1/2048-(1+1/2+1/4+1/8+…+1/1024)

S - ½ S=1-1/2048

           =2047/2048

        S=2047/2048:1/2

           =1,999023438

b)            Giải

   Khoảng cách : 1

   Số số hạng là :

      (100-1):1+1=100(số)

  Tổng các số là :

       (100+1)x100:2=5050

            Đáp số 5050

c)              Giải

     Khoảng cách : 1.1

   Số số hạng là:

       (99,100-1,2):1.1+1=90(số)

  Tổng các số là :

         (99,100+1,2)x90 :2=4513,5

              Đáp số 4513,5

a) Mình có cách khác nha : 

Ta có \(S=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{1024}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+......+\frac{1}{512}\)

\(\Rightarrow2S-S=2-\frac{1}{1024}\)

\(\Rightarrow S=\frac{2047}{1024}\)

\(B=1+2+3+....+200\)

\(2B=1+2+....+200+1+2+...+200\)

\(2B=\left(1+200\right)+\left(2+199\right)+....+\left(1+200\right)\)

\(2B=201+201+....+201\)

\(\Rightarrow B=\frac{200\cdot201}{2}=20100\)

\(S=1.2+2.3+3.4+..+99.100\)

\(3S=1.2.3+2.3.3+3.4.3+....+99.100.3\)

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

\(3S=1.2.3+2.3.4+3.4.5+....+99.100.101\)

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