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

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 \(S\left(100\right)=\frac{99.100.101}{3}=333300\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}=\frac{5}{6}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)

ui cí này e chưa học

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}\)

\(=\frac{5}{6}\)

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

\(A=1.2+2.3+3.4+...+n.\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.4+3.4.3+...+3n.\left(n+1\right)\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n.\left(n+1\right).\left(n+2\right)-\left(n-1\right)n.\left(n+1\right)\)

\(3A=n.\left(n+1\right).\left(n+2\right)\)

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

Vậy \(A=\frac{n.\left(n+1\right).\left(n+2\right)}{3}.\)

Chúc em học tốt!

3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3

=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]

=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]

=n.(n+1).(n+2)

=>S=[n.(n+1).(n+2)] /3

3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3

=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]

=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]

=n.(n+1).(n+2)

=>S=[n.(n+1).(n+2)] /3

ko chắc chắn lắm

3A=1.2.3+2.3.3+3.4.3+4.5.3+.....+9.10.3

3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+.....+9.10.(11-8)

3A=1.2.3-1.2.0+2.3.4-1.2.3+.....+9.10.11-9.10.8

3A=9.10.11

A=(9.10.11):3

A=330

CHẮC CHẮN 100% LÀ ĐÚNG

\(\text{Ta có: A = 1.2+2.3+3.4+4.5+...+99.100 }\)

=>  3A = 3.(1.2+2.3+3.4+4.5+...+99.100)

=>  3A = 1.2.(3 - 0) +2.3.(4 - 1) + 3.4.(5-2) + ........ + 99.100.(101 - 98)

=>  3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .......... + 99.100.101

=>  3A = 99.100.101

\(\Rightarrow A=\frac{99.100.101}{3}=333300\)

k mình nếu đúng OK

Dãy số trên có số lượng các số là :

       (99,100 - 1,2) : 1,1 + 1 = 90 (số)

A = (1,2 + 99,100) x 90 : 2 = 4513,5

             Đáp số : A = 4513,5.

Áp dụng công thức ta có :

\(A=1.2+2.3+3.4+...+99.100=\frac{99.100.101}{3}=333300\)

A=1.2+2.3+3.4+4.5+.....+98.99+99.100 Rút gọn đi ta còn:

A=1+100

=>A=101

Ta có : A = 1.2 + 2.3 + 3.4 + ....... + 99.100

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101

=> 3A = 99.100.101

=> A = 99.100.101/3

=> A = 333300