Mình nghĩ là bạn chép nhầm đề vì nếu là vô số số 1 thì không thể tính được. Đề đúng phải là:

Cho \(A=\frac{2016^2+1^2}{2016.1}+\frac{2015^2+2^2}{2015.2}+...+\frac{1009^2+1008^2}{1009.1008}\); \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\)

Tính \(\frac{A}{B}\)

Ta có: \(A=\frac{2016^2+1^2}{2016.1}+\frac{2015^2+2^2}{2015.2}+...+\frac{1009^2+1008^2}{1009.1008}\)

\(=\frac{2016}{1}+\frac{1}{2016}+\frac{2015}{2}+\frac{2}{2015}+...+\frac{1009}{1008}+\frac{1008}{1009}\)

\(=\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}\)

\(=1+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)\)

\(=1+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}\)

\(=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=2017\)

Xem kỹ là số

\(B=\frac{1+1+...+1}{2+3+...+2016}\) hay \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\) nhé b

b) B = 2² + 4² + 6² + ... + 98² 

 \(\frac{1}{4}B=1^2+2^2+3^2+...+49^2\) 

\(\frac{1}{4}B=1+2\left(1+1\right)+3\left(2+1\right)+...+49\left(48+1\right)\) 

\(\frac{1}{4}B=1+2+1.2+2.3+3+...+48.49+49\) 

\(\frac{1}{4}B=\left(1+2+3+...+49\right)+\left(1.2+2.3+...+48.49\right)\) 

đặt A = 1.2 + 2.3 +...+ 48.49 ta có:

A = 1.2 + 2.3 +...+ 48.49

3A = 1.2.3 + 2.3.( 4 - 1) + ... + 48.49.( 50 - 47 )

3A = 1.2.3 + 2.3.4 - 1.2.3 +...+ 48.49.50 - 47.48.49

3A = 48.49.50

A = \(\frac{48.49.50}{3}=39200\)  

thay A = 39200 vào \(\frac{1}{4}B\) ta có:

\(\frac{1}{4}B=\left(1+2+3+...+49\right)+39200\) 

\(\frac{1}{4}B=1225+39200\)

 \(\frac{1}{4}B=40425\) 

B = 40425.4

B = 161700

vậy B = 161700

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

3A=99.100.101-0

3A=999900

A=999900:3

A=333300

Ta có \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)

\(\Rightarrow A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=2\left(\frac{2016}{2017}\right)\)

\(\Rightarrow A=\frac{4032}{2017}\)

Ta có:\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2016\cdot2017}\)

\(=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+....+\frac{2}{2016}-\frac{2}{2017}\)

\(=\frac{2}{1}-\frac{2}{2017}=2-\frac{2}{2017}=\frac{4034}{2017}-\frac{2}{2017}=\frac{4032}{2017}\)

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 + 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 :>