tính D = 4.5\(^{100}\) .(\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.............+\dfrac{1}{5^{100}}\) )+1
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15 tháng 4 2017
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=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
24 tháng 4 2017
\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)\(\)
\(2P-P=1-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}}{2^{100}}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}-1}{2^{100}}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
6 tháng 5 2022
Đặt biểu thức trong ngoặc đơn là B
\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}+\dfrac{1}{5^{99}}\)
\(\Rightarrow4B=5B-B=1-\dfrac{1}{5^{100}}\Rightarrow B=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)
\(\Rightarrow A=4.5^{100}.\dfrac{1}{4}\left(\dfrac{5^{100}-1}{5^{100}}\right)+1=\)
\(=5^{100}\)
11 tháng 2 2018
\(5D=1+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+...+\dfrac{1}{6.5^{99}}\)
\(6D=\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{6.5^{100}}\)
\(D=\dfrac{\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{36.5^{100}}}{6}\)
MH
17 tháng 10 2023
\(A=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-...-\dfrac{1}{5^{99}}+\dfrac{1}{5^{100}}\)
\(=-\dfrac{1}{5}\left(1-\dfrac{1}{5}\right)-\dfrac{1}{5^3}\left(1-\dfrac{1}{5}\right)-...-\dfrac{1}{5^{99}}\left(1-\dfrac{1}{5}\right)\)
\(=\left(1-\dfrac{1}{5}\right)\left(-\dfrac{1}{5}-\dfrac{1}{5^3}-...-\dfrac{1}{5^{99}}\right)\)
\(=\left(\dfrac{1}{5}-1\right)\left(\dfrac{1}{5}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\right)\)
Mặt khác:
\(F=\dfrac{1}{5}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\)
\(25F=5+\dfrac{1}{5}+...+\dfrac{1}{5^{97}}\)
\(25F-F=5-\dfrac{1}{5^{99}}\)
\(F=\dfrac{5-\dfrac{1}{5^{99}}}{24}\)
\(\Rightarrow A=\left(\dfrac{1}{5}-1\right).F\)
\(=\dfrac{-4}{5}.\dfrac{5-\dfrac{1}{5^{99}}}{24}=\dfrac{\dfrac{1}{5^{99}}-5}{5.6}=\dfrac{\dfrac{1}{5^{100}}-1}{6}\)
NA
14 tháng 4 2022
Đặt `B=1/5+1/5^{2}+1/5^{3}+...+1/5^{101}`
`<=>5B=1+1/5+1/5^{2}+...+1/5^{100}`
`<=>5B-B=(1+1/5+1/5^{2}+...+1/5^{100})-(1/5+1/5^{2}+...+1/5^{101})`
`<=>5B-B=1+1/5+1/5^{2}+...+1/5^{100}-1/5-1/5^{2}-...-1/5^{101}`
`<=>4B=1-1/5^{101}`
`<=>B=(1-1/5^{101})/4`
`@Shả`
14 tháng 4 2022
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{101}}\)
\(5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{100}}\)
\(5A-A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{100}}-\dfrac{1}{5}-\dfrac{1}{5^2}-...-\dfrac{1}{5^{101}}=1-\dfrac{1}{5^{101}}\Rightarrow A=\dfrac{1-\dfrac{1}{5^{101}}}{4}\)
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"
