Ta có : \(n^3-n=n\left(n^2-1\right)=\left(n-1\right).n.\left(n+1\right).\\ \)
=> \(n^3=n+\left(n-1\right).n.\left(n+1\right)\left(1\right)\)
Áp dụng (1) và A ta được :
\(A=1+2+1.2.3+3+2.3.4+4+3.4.5+.....+2016+2015.2016.2017\)
\(A=\left(1+2+3+....+2016\right)+\left(1.2.3+2.3.4+3.4.5+....+2015.2016.2017\right).\)
Đặt C = 1 + 2 +3 + .... + 2016 ; B = 1.2.3 + 2.3.4+ 3.4.5+ .....+ 2015.2016.2017.
\(C=1+2+3+...+2016=\frac{\left(2016+1\right)2016}{2}\)
4 B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +....+(2016- 1).2016.(2016+1).4
4B = 1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6 - 2) + .....+ (2016- 1).2016.(2016+1). [(2016+2) - ( 2016 - 2)]
4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 +.....+ (2015-1).2016.(2016+1).(2016+2) - (2016-2).(2016- 1).2016.(2016+1)
=>\(B=\frac{\left(2016-1\right).2016.\left(2016+1\right).\left(2016+2\right)}{4}.\)
\(A=B+C=\frac{2016.\left(2016+1\right)}{2}+\frac{\left(2016-1\right).2016.\left(2016+1\right).\left(2016+2\right)}{4}.\)
\(A=\left(\frac{\left(2016+1\right).2016}{2}\right)^2=\left(2017.1008\right)^2\)
Số quá to. bạn tự viết nhé