A)

dãy trên có số số hạng là:

( 2016,2017 - 1,2 ) : 1,1 + 1 = 18339,1 chữ số

A là:

( 2016,2017 + 1,2 ) x 18339,1 : 2 = 18560918,35

B)

dãy trên có số số hạng là:

( 2016,2018 - 1,3 ) : 1,1 + 1 = 1832,73 chữ số

B là:

( 2016,2018 + 1,3 ) x 1832,73 : 2 = 1848768,04

\(6x-302=2^3.5\)

\(6x-302=8.5\)

\(6x-302=40\)

\(6x=40+302\)

\(6x=342\)

\(\Rightarrow x=57\)

\(6x-302=2^3.5\)

\(\Rightarrow6x-302=8.5\)

\(\Rightarrow6x-302=40\)

\(\Rightarrow6x=40+302=342\)

\(\Rightarrow x=342:6=57\)

Vậy x = 57

6x-302=2^3.5

6x-302=8.5

6x-302=40

6x       =302+40

6x       =342

 x        =342:6

 x        =57

Vậy x=57

A=3/1.2+3/2.3+3/3.4+3/4.5+...+3/2021.2022

A=3(1/1.2+1/2.3+1/3.4+1/4.5+...+1/2021.2022)

A=3(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2021-1/2022)

A=3[1/1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/2021-1/2021)-1/2022]

A=3[1/1+0+0+0+...+0-1/2022

A=3(1/1-1/2022)

A=3(2022/2022-1/2022)

A=3.2021/2022

A=2021/674

Bn Tham Khảo:

https://hoc247.net/hoi-dap/toan-6/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-2016-faq188428.html

\(A=1\cdot2+2\cdot3+...+151\cdot152\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)

\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)

\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)

\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)

\(=151\cdot76+151\cdot7676=1170552\)

\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)

\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)

\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)

\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)

\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)

\(=4\left[506\cdot1013+345990150\right]\)

\(=1386010912\)

\(M=1^2+2^2+...+2024^2\)

\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)

\(=2024\cdot2025\cdot\dfrac{4049}{6}\)

=2765871900

\(N=1^3+2^3+...+100^3\)

\(=\left(1+2+3+...+100\right)^2\)

\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)

\(=\left[50\cdot101\right]^2=5050^2\)

\(Q=1^3+2^3+...+2024^3\)

\(=\left(1+2+3+...+2024\right)^2\)

\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)

\(=\left[1012\left(2024+1\right)\right]^2\)

\(=2049300^2\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{99\cdot101}\)

\(=2\cdot\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)

\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2} \cdot\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}\cdot\frac{100}{101}\)

\(=\frac{50}{101}\)

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}=\frac{2003}{2004}\)

b) Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2003\cdot2005}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(2A=1-\frac{1}{2005}\)

\(2A=\frac{2004}{2005}\)

\(A=\frac{2004}{2005}:2=\frac{2004}{2005}\cdot\frac{1}{2}=\frac{1002}{2005}\)

a)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=\frac{1}{1}-\frac{1}{2004}\)

\(\Rightarrow=\frac{2003}{2004}\)

b)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003+2005}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(=\frac{1}{1}-\frac{1}{2005}\)

\(\Rightarrow=\frac{2004}{2005}\)

\(a,\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2003.2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)

b) Đặt \(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)

\(\Rightarrow2B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)

\(\Rightarrow2B=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Rightarrow2B=1-\frac{1}{2005}\)

\(\Rightarrow2B=\frac{2005}{2005}-\frac{1}{2005}\)

\(\Rightarrow2B=\frac{2004}{2005}\)

\(\Rightarrow B=\frac{2004}{2005}:2=\frac{2004}{2005}.\frac{1}{2}\)

\(\Rightarrow B=\frac{1002}{2005}\)

Vậy...

hok tốt!!