Question

[Lê Hiển Vinh]

Chứng minh rằng với \(n\in N\)* thì:

a, \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

b, \(1^3+2^3+3^3+...+n^3=\left(\frac{n\left(n+1\right)}{2}\right)^2\)

c, \(n+2\left(n-1\right)+3\left(n-2\right)+...+n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\)