Đặt \(A=\frac{3}{2^3}+...+\frac{100}{2^{100}}\)
\(\frac{1}{2}A=\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{99}{2^{100}}+\frac{100}{2^{101}}\)
\(\Rightarrow A-\frac{1}{2}A=\left(\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)-\left(0+\frac{3}{2^4}+...+\frac{99}{2^{100}}\right)-\frac{100}{2^{101}}\)
\(\frac{1}{2}A=\frac{3}{2^3}-\frac{100}{2^{101}}+\left(\frac{1}{2^4}+\frac{1}{2^5}+...+\frac{1}{2^{100}}\right)\)
Đặt \(B=\frac{1}{2^4}+...+\frac{1}{2^{100}}\)
\(2B=\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2B-B=B=\frac{1}{2^3}-\frac{1}{2^{100}}\)
\(\frac{1}{2}A=\frac{3}{2^3}-\frac{100}{2^{101}}+\left(\frac{1}{2^3}-\frac{1}{2^{100}}\right)=\frac{1}{2}-\frac{100}{2^{101}}-\frac{1}{2^{100}}\)
\(A=1-\frac{100}{2^{100}}-\frac{2}{2^{100}}=1-\frac{102}{2^{100}}\)
Tổng đã cho \(=A+1=2-\frac{102}{2^{100}}\)
Đặt A1 = 1/2^1 + 1/2^2 + ... + 1/2^100
A2 = 1/2^2 + 1/2^3 + ... + 1/2^100
A3 = 1/2^3 + 1/2^4 + ... + 1/2^100
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A100 = 1/2^100
A = 1/2^1 + 2/2^2 + 3/2^3 + 4/2^4 + ... + 100/2^100 =
= (1/2^1+1/2^2 +...+ 1/2^100) + (1/2^2+1/2^3 +...+ 1/2^100) + (1/2^3+1/2^4 +...+ 1/2^100) + ... + (1/2^100) = A1 + A2 + A3 + ... + A100
2^101 A1 = 2^100 + 2^99 + 2^98 + ... + 2 (1)
2^100 A1 = 2^99 + 2^98 + 2^97 + ... + 1 (2)
(2) trừ (1) ---> 2^100 A1 = 2^100 - 1 ---> A1 = (2^100 - 1) / 2^100 = 1 - 1/2^100
Tương tự
2^101 A2 = 2^99 + 2^98 + 2^97 +...+ 2 (3)
2^100 A2 = 2^98 + 2^97 + 2^96 +...+ 1 (4)
(4) trừ (3) ---> 2^100 A2 = 2^99 - 1 ---> A2 = (2^99 - 1) / 2^100 = 1/2 - 1/2^100
Tương tự
A3 = 1/4 - 1/2^100 = 1/2^2 - 1/2^100
A4 = 1/2^3 - 1/2^100
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A100 = 1/2^99 - 1/2^100
Vậy A = A1 + A2 + A3 +...+ A100 = (1 + 1/2 + 1/2^2 + ... + 1/2^99) - 100/2^100
= 2 A1 - 100/2^100 = 2 - 2/2^100 - 100/2^100 = 2 - 51/2^99
Nguồn: TÝnh: A = 1 + 3/2^3 + 4/2^4 +5/2^5 + ....+100/2^100? Ai tra loi dung cho 5 sao? | Yahoo Hỏi & Đáp
