Question

tính tổng : 1² + 2² + 3² +...+ n²

Answer 1

Đặt A =  1² + 2² + 3² +...+ n²

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

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

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

 Đặt  B = 1+ 2+ 3 + ...+ n = (n +1).n/ 2 

Đặt C = 1.2+ 2.3 + 3.4 + ...+ (n-1).n

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

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

3C = [1.2.3 + 2.3.4 + ....+ (n-1).n.(n +1)] - (1.2.3 + 2.3.4 + ... + (n-2)(n -1).n)

3C = (n -1).n (n +1) => C = (n -1).n.(n +1)/ 3

=>  A = (n +1).n/ 2 + (n -1).n(n +1)/3