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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:
\(A=127500-42925=84575\)
Ta thấy:
1/1 + 1/99 = (99+1)/(1.99)=100/(1.99)
1/3 + 1/97 = (97+3)/(3.97)=100/(3.97)
1/5 + 1/95 = (95+5)/(5.95)=100/(3.97)
…
1/97 + 1/3 = (3+97)/(97.3)=100/(97.3)
1/99 + 1/1 = (1+99)/(99.1)=100/(99.1)
=>
1/(1.99)=(1/1+1/99)/100
1/(3.97)=(1/3+1/97)/100
…
1/(99.1)=(1/99+1/1)/100
------------------------------ cộng 2 vế của các đẳng thức trên. Ta được đẳng thức:
1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 )
=[(1/1+1/99)+(1/3+1/99)+…+(1/99+1/1)]/1...
=2(1+1/3+1/5+1/7…+1/99]/100
=(1+1/3+1/5+1/7…+1/99]/50
Vậy:
A=(1+1/3+1/5+1/7+...+1/97+1/99) / [ 1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) ]
A=(1+1/3+1/5+1/7+...+1/97+1/99)/[(1+1/3...
A=50.
ko chắc nhé
Cho A=(1+1/3+1/5+...+1/97+1/99)/(1/1*99+1/3*97+1/5*95+...+1/49*51) rut gon ta duoc A=3
1/ A= 1.(100-1)+2(100-2)+3.(100-3)+...+49.(100-51)+50.(100-50)
= 1.100-1+2.100 - 2.2 + 3.100 -3.3 + ...+49.100 - 49.51 + 50.100 - 50.50
= 100( 1 + 2 + 3 + ...+ 50) - ( 1 + 22 + 32 + ... + 502 )
= 127500- 42925
= 84575
2/ A= 1.3 + 5.7 + 9.11+ 13.15 + 17.19 + ... + 97. 101
= 1.3 + 5(6 + 1) +9( 6+ 5) + 13(6+9) + 17(6+13) + ... + 97(95+6)
= 3 + 5.6 + 1.5 + 9.6 + 5.9 + 13.6 + 9.13 + 17.6 + 13.17 + ... + 95.97 + 97.6
= 3 + ( 1.5 + 5.9 + 9.13 + 13.17 + ...+ 95.97) + 6( 5 + 9 + 13 + 17 + ... + 97)
= ...
=\(\frac{509447}{6}\)