Tính tổng S = \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{8}\) + \(\dfrac{4}{16}\) + ... + \(\dfrac{10}{2^{10}}\)
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22 tháng 4 2021
Tìm y:
-y:1/2-5/2=4+1/2
-y:1/2 = 4+1/2+5/2
-y:1/2 = 7
-y = 7.2
y = -14
Vậy y = -14
26 tháng 2 2022
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
Ta có S = \(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...+\dfrac{10}{2^{10}}\)
= \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\)
2S = 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)
2S - S = ( 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)) - ( \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\))
S = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)
Đặt A = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)
2A = 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)
2A - A = ( 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)) - ( 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\))
A = 2 - \(\dfrac{1}{2^9}\)
⇒ S = 2 - \(\dfrac{1}{2^9}\) - \(\dfrac{10}{2^{10}}\) = \(\dfrac{2^{11}}{2^{10}}-\dfrac{2}{2^{10}}-\dfrac{10}{2^{10}}=\dfrac{2^2\left(2^9-3\right)}{2^{10}}=\dfrac{2^9-3}{2^8}\)
Vậy S = \(\dfrac{2^9-3}{2^8}\)