Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100

3A= 3.(1.2 + 2.3 + 3.4 + ..... +99.100)

3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)

 3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4  + .....+99.100.101

3A=99.100.101

A=99.100.101/3=333300

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3

     = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101

     = 99.100.101

A=333300

B= (1.2 + 2.3 + 3.4 + ... + 100.101) - (1 + 2 + 3+ 4 + ... + 100)

  = 333300 + 10100 - 5050

  = 333300 + 5050

  = 338350

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

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

<=> 3A =1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

<=> 3A =1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

<=> 3A = 99.100.101 = 999900

=> S = 333300

\(A=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=9\left(1-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4

==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)

==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11

==> 4C= 8.9.10.11=7920

==> C= 7920 :4=1980

a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3

               3A=1.2.(3-0) + 2.3.(4-1)+ 3.4.(5-2)+ ... + 99.100.( 101-98)

               3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)

               3A= 99.100.101 - 0.1.2

               3A= 999900 - 0

               3A= 999900

    ==> A= 999900 : 3

   ==> A= 333300

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9\times\frac{99}{100}\)

\(A=\frac{891}{100}\) hoặc =8,91

A=9/1.2+9/2.3+9/3.4+...+9/98.99+9/99.100

A=9.(1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100)

A=9.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)

A=9.(1/1-1/100)

A=9.99/100

A=891/100

A=8+91/100 ( viết dưới dạng hỗn số )

Vậy A=8+91/100

Nkớ k cho mink đó nha !!!

C=1.2+2.3+...+99.100

3C=1.2.3+2.3.3+...+99.100.3

3C=1.2(3-0)+2.3(4-1)+...+99.100(101-98)

C=99.100.101 phần 3

C=333 300

D=2²+4²+6²+333 300

D=4+16+36+333 300

D=20+36+333 300

D=56+333 300

D=333 356

Tích cho mình hai câu nhà, thank

a) \(C=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(\Rightarrow3C=99.100.101\)

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

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

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

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

\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)

\(P=1+0+0+....+0+\dfrac{-1}{100}\)

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

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

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

=1-1/100

=99/100

A = 1.2 + 2.3 + ... + 99.100

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

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

3A = 99.100.101

3A = 999900

A = 333300

C = 1.2.3 + 2.3.4 + ... + 49.50.51

4C = 1.2.3.4 + 2.3.4.(4-1) + ... + 49.50.51.(52-48)

4c = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 49.50.51.52 - 48.49.50.51

4C = 49.50.51.52

4C = 6497400

C = 1624350

Ta có :

a=1.2+2.3+3.4+...+99.100

3a=1.2.3+2.3.3+3.4.3+...+99.100.3

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

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3a=99.100.101

a=\(\frac{99.100.101}{3}\)

a=333300

Tính c làm tương tự

a = 1.2 + 2.3 + 3.4 + ... + 99.100

3a = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3

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

3a = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3a = 99 . 100 . 101

3a = 3 . 33 . 100 . 101

a = 33 . 100 . 101

a = 333300