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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
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
a, \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
\(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)(đpcm)
b, Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{397}{3^{100}}\)
\(A=\frac{3}{4}-\frac{397}{4.3^{100}}< \frac{3}{4}\)(đpcm)
B= (1/2-1/3) + (1/3-1/4) + (1/4-1/5)+...+( 1/99-1/100)
B = (1/2-1/3) + (1/3 - 1/4) + (1/4 - 1/5)+...+ (1/99 + 1/100)
B= 1/2 +1/100=51/100
k mk nhóe
sai thì chỉ mk nhoa
a)A=1/51+1/52+...+1/100
=>A>1/100+1/100+...+1/100
=>A>50/100(vì có 50 số hạng)
=> A>1/2
b)Ta có:
B=1/2.3+1/3.4+...+1/99.100
=> B=1/2-1/3+1/3-1/4+...+1/99-1/100
=> B=1/2-1/100
Mà 1/100>0
=> B<1/2
=> B<1/2<A
=>B<A