a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

! là j z

\("!"\)  là giai thừa đó bạn ạ .

\(VD:\)  \(3!=1.2.3=6\)

          \(4!=1.2.3.4=24\)

a) 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 nên 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100 < 1 (ĐPCM)

a)1-1/2+1/2-1/3+1/3-1/4+......+1/99-1/100

1-1/100=99/100<1

cho mk nha ^^

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

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

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

Vì \(\frac{1}{100}>0\Rightarrow1-\frac{1}{100}< 1\)

\(\frac{\Rightarrow1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}< 1\)

c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)

\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)

\(\Leftrightarrow A=33\cdot100\cdot101=333300\)

b) Ta có: \(1+2-3-4+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=-4\cdot25=-100\)

chua biet lam

ta có 1+(1+2)+(1+2+3)+...+(1+2+3+...+100)

=4+(1+3).3/2+9+(1+4).4/2+...+(1+100).100/2

=1/2(1.2+2.3+.....+100.101)

=>1/2.100.101.102

con cái dưới thì bằng 99.100.101

=>F=51/99

ngu rua mà ko biet lam

2/2*1+3/2*2+4/2*3+5/2*4+6/2*5+....101/2*100=1/2*(2*1+3*2+4*3+5*4+...100*101)=

Giúp Mình với

Mi hả Đức ta Gia Huy nè !

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

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

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

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

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

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

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

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

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

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

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

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

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

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