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2P=\(\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{100}}\)
2P=\(1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
2P-P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)\(\)
\(2P-P=1-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}}{2^{100}}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}-1}{2^{100}}\)
a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)
\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)
\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)
\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)
b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
\(B=4.\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)\)
Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)
\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)
\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)
\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
Ta có :
\(D=\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-..........-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\)
\(5D=1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+..........+\dfrac{1}{5^{100}}\)
\(5D+D=\left(1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+.........+\dfrac{1}{5^{100}}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5^2}+..............-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\right)\)\(6D=1-\dfrac{1}{5^{101}}\)
\(D=\dfrac{1-\dfrac{1}{5^{101}}}{6}\)
*Có: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)\(>\dfrac{1}{4}-\dfrac{1}{101}=\dfrac{97}{404}\)\(=\dfrac{970}{4040}\)
Có: \(\dfrac{1}{5}=\dfrac{808}{4040}\)
\(\Rightarrow\dfrac{1}{5}< A\)
*Có: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)\(=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{5}< A< \dfrac{1}{3}\)
Ta có :
+) \(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
+) \(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
.......................
+) \(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+............+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+......+\dfrac{1}{100}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\) \(\left(1\right)\)
Lại có :
+) \(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)
+) \(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)
.....................
+) \(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.........+\dfrac{1}{100.101}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}>\dfrac{1}{5}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\)
Lại có \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
Đặt N=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+......+\dfrac{1}{5^{100}}\)
5N=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+..........+\dfrac{1}{5^{99}}\)
5N-N= \(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.............+\dfrac{1}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+..........+\dfrac{1}{5^{100}}\right)\)
4N=1-\(\dfrac{1}{5^{100}}\) =\(\dfrac{5^{100}-1}{5^{100}}\)
N=\(\dfrac{5^{100}-1}{4.5^{100}}\)
Thay N vào D ,ta có
D= 4.5\(^{100}\).(\(\dfrac{5^{100}-1}{4.5^{100}}\) )+1
D=5\(^{100}\)
Vậy D =5\(^{100}\)
thank nha, "Thiên Nhi"
