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

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

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

\(\Rightarrow3C=99.100.101\)

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

a)đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

A=1.2+2.3+3.4+4.5+...+99.100

=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101 

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300 

b)D=2 +4 +6 +333 300

D=4+16+36+333 300

D=20+36+333 300 

D=56+333 300

D=333 356 

thank

a) C = 1.2 + 3.4 + 4.5 +...+ 99.100

\(\Rightarrow\) C = 1.2+2.3+3.4+4.5+...+99.100

\(\Rightarrow\)3C = 1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

\(\Rightarrow\)3C = 1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

 \(\Rightarrow\)3C = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

 \(\Rightarrow\)3C = 1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101 

\(\Rightarrow\)3C = 99.100.101

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

b) D = 2 + 4 + 6 + 333 300

\(\Rightarrow\)D = 4 + 16 + 36 + 333 300

\(\Rightarrow\)D = 20 + 36 + 333 300 

\(\Rightarrow\)D = 56 + 333 300

\(\Rightarrow\)D = 333 356 

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

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

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

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

3C = 99.100.101

C = 99.100.101 :3

C = 333300

Câu b tự làm

P/S : Hoq chắc :>

Phần b mình chưa nghĩ ra

a) C = 1 . 2 + 2 . 3 + ... + 49 . 50

=> 3C = 1 . 2 . 3 + 2 . 3 . 3 + ... + 49 . 50 . 3

=> 3C = 1 . 2 . ( 3 - 0 ) + 2 . 3 . ( 4 - 1 ) + ... + 49 . 50 . ( 51 - 48 )

=> 3C = 1 . 2 . 3 - 0 . 1 . 2 + 2 . 3 . 4 - 1 . 2 . 3 + ... + 49 . 50 . 51 - 48 . 49 . 50

=> 3C = 49 . 50 . 51

=> C = 49 . 50 . 17 = 41650

\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)

\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)

\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)

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

 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

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

 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

sưả đề \(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)