Tính
V=4.5^100(1/5+1/5^2+1/5^3+.....+1/5^100)+1
Những câu hỏi liên quan
4.5100.(1/5+1/52+1/53+....+1/5100)+1
Đặt:
\(A=4.5^{100}.\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.....+\dfrac{1}{5^{100}}\right)+1\)
\(S=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.....+\dfrac{1}{5^{100}}\)
\(5S=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.....+\dfrac{1}{5^{100}}\right)\)
\(5S=1+\dfrac{1}{5}+\dfrac{1}{5^2}+.....+\dfrac{1}{5^{99}}\)
\(5S-S=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+.....+\dfrac{1}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{100}}\right)\)\(4S=1-5^{100}\Rightarrow S=\dfrac{1-5^{100}}{4}\)
Thay S và A ta có:
\(A=4.5^{100}.\dfrac{1-5^{100}}{4}+1\)
\(A=5^{100}.\left(1-5^{100}\right)+1\)
\(A=5^{100}-5^{200}+1\)
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Tinh
\(V=4.5^{100}\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+....+\frac{1}{5^{100}}\right)+1\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5A-A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)\)
\(4A=1-\frac{1}{5^{100}}\)
\(A=\frac{1-\frac{1}{5^{100}}}{4}\)
\(A=\frac{1}{4}-\frac{1}{5^{100}}:4\)
\(A=\frac{1}{4}-\frac{1}{5^{100}.4}\)
=> \(V=4.5^{100}.\left(\frac{1}{4}-\frac{1}{5^{100}.4}\right)+1\)
\(V=\left(4.5^{100}.\frac{1}{4}-4.5^{100}.\frac{1}{5^{100}.4}\right)+1\)
\(V=\left(5^{100}-1\right)+1\)
\(V=5^{100}\)
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Tính
V=\(4.5^{100}.\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)+1\)1
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+....+\frac{1}{5^{100}}\)
\(5A=5+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}\)
\(5A-A=1-\frac{1}{5^{100}}\)
\(4A=1-\frac{1}{5^{100}}\)
\(A=\frac{1-\frac{1}{5^{100}}}{4}\)
\(A=\frac{1}{4}-\frac{1}{4.5^{100}}\)
\(V=4.5^{100}\left(\frac{1}{4}_{ }-\frac{1}{4.5^{100}}\right)+1\)
\(V=\left(4.5^{100}.\frac{1}{4}-4.5^{100}.\frac{1}{4.5^{100}}\right)+1\)
\(V=\left(5^{100}-1\right)+1\)
\(V=5^{100}\)
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V = \(4.5^{100}.\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\right)+1\)
GIÚP MK VỚI
tính D = 4.5\(^{100}\) .(\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.............+\dfrac{1}{5^{100}}\) )+1
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\(^{100}\)
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Tính tổng sau:
4.5100(1 phần 52+1 phần 53 +1 phần 54+...+1 phần 5100) +1
Tính \(V=4.5^{100}.\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{100}}\right)+1\)
Tính :
C = \(4.5^{100}\left(\frac{1}{5}+\frac{1}{5^2}+....+\frac{5}{5^{100}}\right)+1\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{1024}\right)\)
\(B=4.5^{100}.\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)+1\)