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\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
\(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
Trừ dưới cho trên:
\(4A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
\(20A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)
Lại trừ dưới cho trên:
\(16A=1-\frac{100}{5^{99}}+\frac{99}{5^{100}}\)
\(\Rightarrow A=\frac{1}{16}-\frac{1}{16.5^{99}}\left(100-\frac{99}{5}\right)< \frac{1}{16}\) do \(100-\frac{99}{5}>0\)
Ta có:
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16
\(B=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\)
\(B=\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}\right)+\left(\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\right)\)
\(B=\dfrac{49}{36}+\left(\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\right)\)
\(B=\dfrac{1}{36}+\dfrac{4}{3}+\left(\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\right)\)
\(B>\dfrac{4}{3}\left(1\right)\)
\(\)\(B< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{24.25}\)
\(B< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(B< 2-\dfrac{1}{25}\)
\(B< 2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\dfrac{4}{3}< B< 2\)
\(\rightarrowđpcm\)
De sai o dau phai hok ban. Phien ban xem lai giup.Toi mik giai cho
Ta có 1/2*3=1/2-1/3;
1/3*4=1/3-1/4
......................(tương tự với các số khác)
1/149*150=1/149-1/150
=>A=1/2-1/3+1/3-1/4+1/4-1/5+...-1/149+1/149-1/150=1/2-1/150
A=75/150-1/150=74/150=37/75
Vậy A= 37/75
Đặt A = 1 + 2 + 22 + 23 + ... + 299 + 2100
2A = 2 + 22 + 23 + 24 + ... + 2100 + 2101
2A - A = (2 + 22 + 23 + 24 + ... + 2100 + 2101) - (1 + 2 + 22 + 23 + ... + 299 + 2100)
A = 2101 - 1 (đpcm)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)