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1.99+2.98+3.97+...+98.2+99.1=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+98.99)
=99.(1+2+...+99)-(1.2+2.3+...+98.99)=99.4950-(1.2+2.3+...+98.99)=490050-(1.2+2.3+...+98.99)
đặt A=1.2+2.3+...+98.99
=>3A=1.2.3+2.3.3+...+98.99.3
=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3-1.2.3+2.3.4-2.3.4+...+97.98.99-97.98.99+98.99.100=98.99.100
=>A=98.99.100:3=323400
=>1.99+2.98+3.97+...+98.2+99.1=490050-323400=166650
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=1.99+2.98+3.97+........+97.3+98.2+99.1\)
\(\Rightarrow C=1.99+2.\left(99-1\right)+3\left(99-2\right)+..........+98.\left(99-97\right)+99.\left(99-98\right)\)
\(\Rightarrow C=1.99+2.99-1.2+3.99-2.3+........+98.99-97.98+99.99-98.99\)
\(\Rightarrow C=\left(1.99+2.99+.......+99.99\right)-\left(1.2+2.3+.........+98.99\right)\)
\(\Rightarrow C=490050-\left(1.2+2.3+....+98.99\right)\)
Đặt \(A=1.2+2.3+3.4+........+98.99\)
\(\Rightarrow3A=1.2.3+2.3.3+..........+98.99.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+.....+98.99\left(100-97\right)\)
\(\Rightarrow3A=1.2.3-1.2.3+2.3.4-2.3.4+......+97.97.99-97.98.99+98.99.100\)
\(\Rightarrow3A=98.99.100\)
\(\Rightarrow A=\dfrac{98.99.100}{3}=323400\)
\(\Rightarrow C=490050-323400=166650\)
Vậy \(C=166650\)
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1.1+3.1+5.1+....+99.1
= 1+3+5+...+99
= (99+1) .[(99-1):2+1] : 2
= 2500
mik nha !!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:Xét tử số
TS có 99 tổng,1 có mặt trong 99 tổng,2 có mặt trong 98 tổng,3 có mặt trong 97 tổng,...,99 có mặt trong 1 tổng
Vì thế ta được:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+1.99}\)
\(=\frac{1.99+2.98+3.97+...+99.1}{1.99+2.98+3.97+...+99.1}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{97}+\dfrac{1}{3}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{\left(\dfrac{1}{1\cdot99}+\dfrac{1}{99\cdot1}\right)+\left(\dfrac{1}{97\cdot3}+\dfrac{1}{97\cdot3}\right)+...+\left(\dfrac{1}{51\cdot49}+\dfrac{1}{49\cdot51}\right)}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{97\cdot3}+...+\dfrac{100}{49\cdot51}}{\dfrac{2}{1\cdot99}+\dfrac{2}{97\cdot3}+...+\dfrac{2}{51\cdot49}}\)
\(=\dfrac{100\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}{2\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a bn tự lm nha mk lm đỡ bn phần b
b=1.99+1.98+3.97+...+98.2+99.1
= 1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
= 1.99+2.99-1.2+3.99-2.3+...98.99-97.98+99.99-99.98
=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+...+98.99)
=99.(1+2+3+...+98+99)-(1.2+2.3+...+98.99)
=99.4950-(1.2+2.3+...+98.99)
=490050-(1.2+2.3+...+98.99)
b=1.2+2.3+...+98.99
3b=1.2.3+2.3.3+...+98.99.3
3b=1.2.3+2.3.(4-1)+...+98.99.(100-97)
3b=1.2.3+2.3.4+2.3.1+...+98.99.100+98.00.97
3b=490050-(98.99.100):3
b=490050-323400
b=166650
tk mk nha
=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99(99-98)
=99.(1+2+3+4+...+98+99)-(2+2.3+3.4+...+97.98+98.99)
=99.(1+99).99/2-98.99.100/3
=99.50.99-98.33.100
=490050-323400
=166650