tinh
A = \(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\)
K
Khách
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28 tháng 9 2015
\(2.A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
=> 2.A - A = \(\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)
=> A = \(\left(2+\frac{3}{2^2}-1-\frac{100}{2^{100}}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{2^4}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)\)
A = \(1+\frac{3}{2^2}-\frac{100}{2^{100}}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}=\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)+\frac{2}{2^2}-\frac{100}{2^{100}}\)
Tính B = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
2.B = \(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\) => 2.B - B = \(1+\frac{1}{2}-\frac{1}{2^{99}}\)=> B = \(\frac{3}{2}-\frac{1}{2^{99}}\)
Vậy A = \(\frac{3}{2}-\frac{1}{2^{99}}+\frac{2}{2^2}-\frac{100}{2^{100}}=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}=2=\frac{2^{101}-102}{2^{100}}\)
11 tháng 9 2017
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
17 tháng 1 2016
\(S=1+\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(S=1+\frac{1+2+3+4+...+101}{2}\)
\(S=1+\frac{10201}{2}=...\)
tick cho mink nha!
TL
1
22 tháng 7 2016
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5A-A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)\)
\(4A=1-\frac{1}{5^{100}}\)
\(A=\frac{1-\frac{1}{5^{100}}}{4}\)
\(A=\frac{1}{4}-\frac{1}{5^{100}}:4\)
\(A=\frac{1}{4}-\frac{1}{5^{100}.4}\)
=> \(V=4.5^{100}.\left(\frac{1}{4}-\frac{1}{5^{100}.4}\right)+1\)
\(V=\left(4.5^{100}.\frac{1}{4}-4.5^{100}.\frac{1}{5^{100}.4}\right)+1\)
\(V=\left(5^{100}-1\right)+1\)
\(V=5^{100}\)