Question

Cho N thuộc N* . Tính tổng

n^2 + ( n + 2 )^2 + ( n +4 )^2 +.........+( n + 100 )^2

Answer 1

Đặt A = biểu thức cần tính. Ta có:

(n+2)²=(n+2)(n+3-1)=(n+2)(n+3)-(n+2)

(n+4)²=(n+4)(n+5-1)=(n+4)(n+5)-(n+4)

....

(n+100)²=(n+100)(n+101-1)=(n+100)(n+101)-(n+100)

A=n²+(n+2)(n+3)-(n+2)+(n+4)(n+5)-(n+4)+...(n+100)(n+101)-(n+100)

=> A=n²+[(n+2)(n+3)+(n+4)(n+5)+...+(n+100)(n+101)]-(50n+2+4+...+100)

=> A=n²​-(50n+2550)+[(n+2)(n+3)+(n+4)(n+5)+...+(n+100)(n+101)]

=> \(A=n^2-50\left(n+51\right)+\frac{\left(n+100\right)\left(n+101\right)\left(n+102\right)}{3}\)