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![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}\) nha
k cho mk nhé
đặt tổng trên là A
ta có:
3A=1.2.3+2.3.3+...+n.(n+1).3
3A=1.2.3+2.3.(4-1)+...+n(n+1)[(n+2)-(n-1)]
3A=1.2.3+2.3.4-1.2.3+...+n(n+1)(n+2)-(n-1)n(n+1)
3A=n(n+1)(n+2)
A=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
D = 1.2 + 2.3+ 3.4 +...+ 99.100
=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
=99.100.101-0.1.2
=99.100.101
=999900
=>D=999900:3=333300
Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)
=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]
=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)
=n.(n+1).(n+2)-0.1.2
=n.(n+1)(n+2)
=>Dn=n.(n+1)(n+2):3
=>điều cần chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)
3C=2014.2015.2016
C=2014.2015.2016:3
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1.22+2.32+..............+(n-1).n2
A=1.2.2+2.3.3+.......+(n-1).n.n
A=1.2.(3-1)+2.3.(4-1)+.........+(n-1).n.(n+1-1)
A=1.2.3-1.2+2.3.4-2.3+..........+(n-1).n.(n+1)-(n-1).n
A=[1.2.3+2.3.4+.........+(n-1).n.(n+1)]-[1.2+2.3+............+(n-1).n)
Bạn tự làm tiếp nhá
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Nguyễn Thị Thu Chi
- S=1.2+2.3+3.4+.............+n(n+1)
S =1(1+1) + 2(2+1) + 3(3+1) +...+n(n+1)
S =(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n)
ta có các công thức:
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6
1 + 2 + 3 + ...+ n = n(n+1)/2
thay vào ta có:
S = n(n+1)(2n+1)/6 + n(n+1)/2
=n(n+1)/2[(2n+1)/3 + 1]
=n(n+1)(n+2)/3
ko chắc chắn lắm
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101
=> 3S = 99.100.101
=> S = \(\frac{99.100.101}{3}=333300\)
ta xét
\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)
\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)
\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)
Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)