ta có CT:
\(\left(n-1\right)n\left(n+1\right)=\left(n^2-n\right)\left(n+1\right)=\text{ }\left(n^2-n\right)n+n^2-n\)
\(=n^3-n^2+n^2-n=n^3-n\)
\(\Leftrightarrow n^3=\left(n-1\right)n\left(n+1\right)n\)
Áp dụng vào A ta được
\(A=\left(1-1\right).1.\left(1+1\right)+1+\left(2-1\right).2.\left(2+1\right)+2+....+\)
\(\left(99-1\right).99.\left(99+1\right)+99+\left(100-1\right).100.\left(100+1\right)+100\)
\(=1+2+1.2.3+3+2.3.4+...+100+99.100.101\)
\(=\left(1+2+3+...+100\right)+\left(1.2.3.+2.3.4+...+99.100.101\right)\)
\(=5050+25497450=25502500\)
Ta có: (n - 1)n(n + 1) = n3 - n => n3 = n + (n - 1)n(n + 1)
Áp dụng vào ta được:
Nhận xét: 13 = 1
23 = 2 + 1.2.3
33 = 3 + 2.3.4
.........
993 = 99 + 98.99.100
1003 = 100 + 99.100.101
=> A = 1 + 2 + 1.2.3 + 3 + 2.3.4 +....+ 99 + 98.99.100 + 100 + 99.100.101
= (1 + 2 + 3 +....+ 99 + 100) + (1.2.3 + 2.3.4 +....+ 98.99.100 + 99.100.101)
Đặt B = 1 + 2 + 3 +...+ 99 + 100
= (1 + 100).100 : 2
= 101.50
= 5050
Đặt C = 1.2.3 + 2.3.4 +....+ 98.99.100 + 99.100.101
4C = 1.2.3.(4 - 0) + 2.3.4.(5 - 1) +...+ 98.99.100.(101 - 97) + 99.100.101.(102 - 98)
4C = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 +...+ 98.99.100.101 - 97.98.99.100 + 99.100.101.102 - 98.99.100.101
4C = (1.2.3.4 + 2.3.4.5 +....+ 98.99.100.101 + 99.100.101.102) - (0.1.2.3 + 1.2.3.4 + .... + 97.98.99.100 + 98.99.100.101)
4C = 99.100.101.102 - 0.1.2.3
4C = 99.100.101.102
C = \(\dfrac{99.100.101.102}{4}\)
C = 25497450
=> A = B + C = 5050 + 25497450 = 25502500
Vậy A = 25502500