\(A=\left(1+3+5+...+2002\right).\left(135135.137-135.137137\right)\)
Đặt : \(C=135135.137-135.137137\)
\(C=\left(135.1001\right).137-135.137137\)
\(C=135.\left(137.1001\right)-135.137137\)
\(C=135.137137-135.137137\)
\(C=0\)
Thay vào ta có :
\(A=\left(1+3+5+...+2002\right).0\)
\(A=0\)
Vậy A = 0
\(B=1.2+2.3+3.4+...+99.100\)
\(3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(3B=999900\)
\(\Rightarrow B=333300\)
Vậy B = 333300
Gọi A là biểu thức ta có:
B = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
B . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
B . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
B. 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
B . 3 = 99.100.101
B = 99.100.101 : 3
B = 33.100.101
B = 333 300
