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=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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Khi đó:
$1+\frac{1}{1.3}=\frac{2^2}{1.3}$
$1+\frac{1}{2.4}=\frac{3^2}{2.4}$
.........
$1+\frac{1}{99.101}=\frac{100^2}{99.101}$
Khi đó:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)
6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101
6A=3+99x101x103=1019703
vậy = 1019703
nếu sai chỗ nào thì sửa hộ mk vs
![](https://rs.olm.vn/images/avt/0.png?1311)
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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
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1x3 +3x5 +5x7 +....+99x101
6A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)
6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101
6A=3+99x101x103=1019703
\(S=1.3+2.4+3.5+...+99.101\)
\(\Rightarrow S=1\left(2+1\right)+2\left(3+1\right)+...+99\left(100+1\right)\)
\(\Rightarrow S=\left(1.2+2.3+...+99.100\right)+\left(1+2+3+...+99\right)\)
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