Question

[conan]

1.2+2.3+3.4+...+n.(n+1)

Answer 1

[Trương Ngọc Đức]

gọi a₁=1.2=>3a₁=1.2.3=>3a₁=1.2.3-0.1.2

a₂=2.3=>3a₂=2.3.3=>3a₁=2.3.4-1.2.3

tương tự .....

đến a_n-1=(n-1)n=>3a_n-1=3(n-1)n =>a_n-1=(n-1)n(n+1)-(n-2)(n-1)n

a_n=n.(n+1)=>3a_n=3.n(n+1)=>3a_n=n(n+1)(n+2)-(n-1)n(n+1)

cộng từng vế các đẳng thức ta được:

3a₁+3a₂+....+3a_n-1+3a_n=1.2.3-0.1.2+2.3.4-1.2.3+....+n(n+1)(n+2)-(n-1)n(n+1)

3(a₁+a₂+...+a_n-1+a_n)=n(n+1)(n+2)

3(1.2+2.3+3.4+...+n(n+1)=n(n+1)(n+2)

=>1.2+2.3+3.4+...+n.(n+1)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Answer 2

[Trần Hoài Bão]

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

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+....+n\left(n+1\right).3\)\(=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+n\left(n-1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

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

\(=n\left(n+1\right)\left(n+2\right)\)

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