Câu hỏi của Lưới Hái Tử Thần

Question

Tính:

$1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)$

Ngô Tấn Đạt · Accepted Answer

$A=1.2+2.3+...+n\left(n+1\right)\
3.A=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\
3.A=n\left(n+1\right)\left(n+2\right)\
\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}$

Chúc bạn học tốt !!!!!!!!!Câu trả lời: $A=1.2+2.3+...+n\left(n+1\right)\
3.A=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\
3.A=n\left(n+1\right)\left(n+2\right)\
\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}$

Chúc bạn học tốt !!!!!!!!!

Khôngg Tồnn Tạii · Answer

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

$\Rightarrow3A=3.1.2+3.2.3+3.3.4+...+3.n.\left(n+1\right)$

Ta có: $3.1.2=1.2.3-0.1.2$

$3.2.3=2.3.4-1.2.3$

$3.3.4=3.4.5-2.3.4$

.....

$3.n.\left(n+1\right)=n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)$$\Rightarrow A=\frac{1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-....+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)}{3}$

$\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)-0.1.2}{3}$

$\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}$Câu trả lời: Đặt $A=1.2+2.3+3.4+...+n.\left(n+1\right)$