Question

chứng minh: \(1^2+2^2+3^2+4^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Answer 1

Đặt A = 1² + 2² + 3² + 4² + ... + n²

A = 1 + (1 + 1).2 + (1 + 2).3 + (1 + 3).4 + ... + (1 + n - 1).n

A = 1 + 1.2 + 2 + 3 + 2.3 + 4 + 3.4 + ... + n + (n - 1).n

A = [1.2 + 2.3 + 3.4 + ... + (n - 1).n] + (1 + 2 + 3 + 4 + ... + n)

A = [1.2 + 2.3 + 3.4 + ... + (n - 1).n] + \(\frac{n.\left(n+1\right)}{2}\)

Đặt B = 1.2 + 2.3 + 3.4 + ... + (n - 1).n

3B = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + (n - 1).n.[(n + 1) - (n - 2)]

3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + (n - 1).n.(n + 1) - (n - 2).(n - 1).n

3B = (n - 1).n.(n + 1)

\(B=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)

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

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

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