Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}\) 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 \(S\left(100\right)=\frac{99.100.101}{3}=333300\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
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
- 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
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
Áp dụng công thức ta có :
\(A=1.2+2.3+3.4+...+99.100=\frac{99.100.101}{3}=333300\)
Lời giải:
$A=1.2+2.3+3.4+...+8.9+9.10$
$3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+....+8.9(10-7)+9.10(11-8)$
$=(1.2.3+2.3.4+3.4.5+...+8.9.10+9.10.11)-(0.1.2+1.2.3+2.3.4+...+8.9.10)$
$=9.10.11$
$\Rightarrow A=\frac{9.10.11}{3}=330$
A=1.2+2.3+...+n(n+1)
3A=1.2.3+2.3.3+....+3n(n+1)
3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)
3A=n(n+1)(n+2)
A=n(n+1)(n+2)/3 (đpcm)
457896325544552445552144521456255222222222222222
\(A=1.2+2.3+3.4+...+n\left(n+1\right)\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
Suy ra \(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)