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Mk sửa lại đề xíu, có lẽ bn chép sai ở phân số cuối của D phải là 1/101
C = 1002+12/100.1 + 992+22/99.2 + ... + 512+502/51.50
C = 1002/100.1 + 12/100.1 + 992/99.2 + 22/99.2 + ... + 512/51.50 + 502/51.50
C = 100/1 + 1/100 + 99/2 + 2/99 + ... + 51/50 + 50/51
C = 100/1 + 99/2 + 98/3 + ... + 51/50 + 50/51 + ... + 1/100
C = (1 + 1 + ... + 1) + 99/2 + 98/3 + ... + 1/100
100 số 1
C = (99/2 + 1) + (98/3 + 1) + ... + (1/100 + 1) + 1
C = 101/2 + 101/3 + ... + 101/100 + 101/101
C = 101.(1/2 + 1/3 + ... + 1/100 + 1/101)
=> C : D = 101
Mk sửa lại đề xíu, có lẽ bn chép sai ở phân số cuối của D phải là 1/101
C = 1002+12/100.1 + 992+22/99.2 + ... + 512+502/51.50
C = 1002/100.1 + 12/100.1 + 992/99.2 + 22/99.2 + ... + 512/51.50 + 502/51.50
C = 100/1 + 1/100 + 99/2 + 2/99 + ... + 51/50 + 50/51
C = 100/1 + 99/2 + 98/3 + ... + 51/50 + 50/51 + ... + 1/100
C = (1 + 1 + ... + 1) + 99/2 + 98/3 + ... + 1/100
100 số 1
C = (99/2 + 1) + (98/3 + 1) + ... + (1/100 + 1) + 1
C = 101/2 + 101/3 + ... + 101/100 + 101/101
C = 101.(1/2 + 1/3 + ... + 1/100 + 1/101)
=> C : D = 101
Đặt \(A=\frac{\frac{2000}{11}+\frac{2000}{12}+...+\frac{2000}{100}}{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}}\)
\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1}\)
\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}\)
\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{100.\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)}\)
\(\Rightarrow A=\frac{20.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}}\)
\(\Rightarrow A=\frac{\frac{20}{11}+\frac{20}{12}+..+\frac{20}{100}}{\frac{1}{99}+\frac{1}{98}+..+\frac{1}{2}+\frac{1}{100}}\)
a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
=\(1-\frac{1}{100!}< 1\)
\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
M=100
Xét tử N
92-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=(1+1+1+...+1)-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=1-(1/9)+1-(2/10)+1-(3/11)+......+1-(90/98)+1-(91/99)+1-(92/100)
=(8/9)+(8/10)+(8/11)+ ...+ (8/98)+(8/99)+(8/100)
=8.[(1/9)+(1/10)+(1/11)+...+(1/98)+(1/99)+(1/100)]
=40[(1/45)+(1/50)+(1/55)+...+(1/495)+(1/500)]
=>N=40
=>M/N=5/2
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}\)
\(B=\frac{1}{100}\)
