A+B = (1.99+2.98+3.97+...+99.1)+(1.101+2.102+3.103+...+99.199)

A+B = (1.99+1.101)+(2.98+2.102)+(3.97+3.103)+...+(99.1+99.199)

A+B = 1(99+101) + 2(98+102) + 3(97.103)+...+99(1+199)

A+B = 1.200 + 2.200 + 3.200 +...+ 99.200

A+B = 200.(1+2+3+...+200)

A+B = 200.4950

A+B = 990000

A+B=(1.99+2.98+...+99.1)+(1.101+2.102+...+99.199)

=(1.99+1.101)+(2.98+2.102)+...+(99.1+99.199)

=1.(99+101)+2.(98+102)+...+99(1+199)

=200+2.200+...+99.200

=200.(1+2+3+4+...+99)

=200.4950

=.....

Đăt A=1.3+5.7+9.11+...+97.99
      B=3.5+7.9+11.13+...+99.101
Ta có: A+B=1.3+3.5+5.7+7.9+...+97.99+99.101
6(A+B)=1.3.6+3.5.6+5.7.6+…+97.99.6+99.101.6
=1.3.(5+1)+3.5.(7−1)+5.7(9−3)+…+97.99(101−95)+99.101.(103−97)
=1.3.5+1.3+3.5.7−1.3.5+5.7.9−3.5.7+…+97.99.101−95.97.99+99.101.103−97.99.101
=3+99.101.103=1029900
⇒A+B=171650
Lại có: 
B−A=3.(5−1)+7.(9−5)+11.(13−9)+...+99.(101−97)
=4.(3+7+11+...+99)
=4.(3+99).252=5100
Suy ra ta có: {A+B=171650B−A=5100⇔ {A=83275B=88375
Vây 1.3+5.7+9.11+...+97.99=83275

b, B= 1.2.3+2.3.4+3.4.5+...+98.99.100

4B = (1.2.3+2.3.4+3.4.5+...+98.99.100).4

4B=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.....+98.99.100.(101-97)

4B= 1.2.3.4-0.1.2.3 + 2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100

4B=98.99.100.101

B=98.99.25.101

B=6999300

a) \(A=1.99+2.98+3.97+...+50.50\)

\(A=1.\left(100-1\right)+2.\left(100-2\right)+3.\left(100-3\right)+...+50.\left(100-50\right)\)

\(A=100.\left(1+2+3+...+50\right)-\left(1^2+2^2+3^2+...+50^2\right)\)

Xét \(1+2^2+3^2+...+50^2\)

\(=1.\left(2-1\right)+2.\left(3-1\right)+3\left(4-1\right)+...+50.\left(51-1\right)\)

\(=\left(1.2+2.3+3.4+...+50.51\right)-\left(1+2+3+...+50\right)\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+...+50.51.\left(52-49\right)}{3}-1275\)

\(=\frac{1.2.3-1.2.3+2.3.4-...-49.50.51+50.51.52}{3}-1275\)

\(=44200-1275\)

\(=42925\)

Thay vào A ta được:

\(A=127500-42925=84575\)

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