Question

[Phạm Kiều Anh]

chứng minh công thức;

\(1^2+2^2+3^2+......+n^2=n\left(n+1\right)\left(2n+1\right)\)

Answer 1

[Nguyễn Hoàng Anh Phong]

1² +2²+3²+...+n²

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

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

Dat A = 1.2+2.3+3.4+...+n.(n+1)

=> 3A = 1.2.3+2.3.3+3.4.3+...+n.(n+1).3

3A = 1.2.3+2.3(4-1)+3.4.(5-2)+...+n.(n+1).(n+2-n+1)

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

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

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

ta co: 1+2+...+n = n.(n+1)/2

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

cop sai de hay sao z bn???

Answer 2

[Khánh Ngọc]

Sửa đề : 1² + 2² + 3² + ... + n² = \(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

VT <=> 1 ( 2 - 1 ) + 2 ( 3 - 1 ) + 3 ( 4 - 1 ) + ... + n [ ( n + 1 ) - 1 ]

= [ 1 . 2 + 2 . 3 + 3 . 4 + ... + n ( n + 1 ) ] - ( 1 + 2 + 3 + 4 + ... + n ) 

Đặt A = 1 . 2 + 2 . 3 + 3 . 4 + ... + n ( n + 1 ) . Ta có :

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 3n ( n + 1 )

=> 3A = 1.2.3 + 2.3 ( 4 - 1 ) + 3.4 ( 5 - 2 ) + ... + n ( n + 1 ) [ ( n + 2 ) - ( n - 1 ) ]

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + n ( n + 1 ) ( n + 2 ) -  ( n - 1 ) n ( n + 1 )

=> 3A = n ( n + 1 ) ( n + 2 )

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

=> VT = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)- ( 1 + 2 + 3 + 4 + ... + n ) 

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

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

\(=\frac{n\left(n+1\right)\left(n+2\right)}{6}=VP\)( Đpcm )