Bài 1:
\(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(3A=n\left(n+1\right)\left(n+2\right)\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Bài 2:
\(B=1^2+2^2+...+n^2\)
\(B=1\left(2-1\right)+2\left(3-1\right)+...+n\left[\left(n+1\right)-1\right]\)
\(B=\left[1\cdot2+2\cdot3+...+n\left(n+1\right)\right]-\left(1+2+...+n\right)\)
\(B=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)
\(B=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
