=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)

=1.2+1+2.3+2+3.4+3+...+100.101+100

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

=333300+5000

=338300

a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)

    Tổng A là: (2020+5)x404:2=409050

b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)

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

          \(=1-\frac{1}{101}=\frac{100}{101}\)

c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)

         \(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)

           \(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)

             \(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)

Vậy .....

A = 5 + 10 + 15 + ... + 2015 + 2020

Số số hạng là : 404

A = 409050

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

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)

\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)

Đáp án =2525 vì câu của cậu có người hỏi rồi

sai rồi

bấm máy tính là ra

Tính nhanh nhé

\(S=1.3+2.4+3.5+...+99.101\)

\(\Rightarrow S=1\left(2+1\right)+2\left(3+1\right)+...+99\left(100+1\right)\)

\(\Rightarrow S=\left(1.2+2.3+...+99.100\right)+\left(1+2+3+...+99\right)\)

Đặt \(A=1.2+2.3+...+99.100\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3S=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

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

Đặt \(B=1+2+3+...+99\)

\(\Rightarrow B=\frac{\left(99+1\right)\left[\left(99-1\right):2+1\right]}{2}\)

\(\Rightarrow B=\frac{100.50}{2}=2500\)

\(\Rightarrow S=A+B=\frac{99.100.101}{3}+2500\)

S = 1 x 3 + 2 x 4 + 3 x 5 + ... + 99 x 101

S = ( 1 x 3 + 3 x 5 + ...+ 99 x 101) +  ( 2 x 4 + ...+ 98 x 100)

Đặt A = 1 x 3 + 3 x 5 + ...+ 99 x 101

=> 6 A = 1 x 3 x 6 + 3 x 5 x 6 + ...+ 99 x 101 x 6

6 A = 1 x 3 x ( 5+1) + 3 x 5 x ( 7-1) + ...+ 99 x 101 x ( 103 - 97)

6A = 1 x 3 x 5 + 1 x 3 + 3 x 5 x 7 - 1 x 3 x 5 + ...+ 99 x 101 x 103 - 97 x 99 x 101

6A = ( 1 x 3 + 1 x 3 x 5 + 3 x 5 x 7 +...+ 99 x 101 x 103) - ( 1 x 3 x 5 + ...+ 97 x 99 x 101)

6A = 1 x  3 + 99 x 101 x 103

\(\Rightarrow A=\frac{1.3+99.101.103}{6}=171650\)

Đặt B = 2 x 4 + ...+ 98 x 100

=> 6B = 2 x 4 x 6 + 4 x 6 x 6 + ...+ 98 x 100 x 6

6B = 2 x 4 x 6 + 4 x 6 x ( 8-2) + ...+ 98 x 100 x ( 102 - 96)

6B = 2 x 4 x 6 + 4 x6 x8 - 2x4x6 + ...+ 98x100x102 - 96x98x100

6B = ( 2 x 4 x 6 + 4 x 6 x 8 +...+98x100x102) - ( 2x4x6+...+96x98x100)

6B = 98 x 100 x 102

\(\Rightarrow B=\frac{98.100.102}{6}=166600\)

Thay A;B vào S, có
S = 171 650 + 166 600

S = 338 250

Hỏi thật hả. 

chịu vì em hok lớp 6