Tính tổng :
\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+..+\dfrac{3}{2^9}\)
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
a. Dãy là tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=\dfrac{1}{10}\end{matrix}\right.\)
Do đó: \(S=\dfrac{u_1}{1-q}=\dfrac{1}{1-\dfrac{1}{10}}=\dfrac{10}{9}\)
b. Tương tự, tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-\dfrac{1}{3}\end{matrix}\right.\) bạn tự ráp công thức
c. \(S=2+S_1\) với \(S_1\) là cấp số nhân lùi vô hạn \(\left\{{}\begin{matrix}u_1=\dfrac{3}{10}\\q=\dfrac{3}{10}\end{matrix}\right.\)
NN
2
10 tháng 5 2017
\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)
\(=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
Đặt \(A=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A=2+1+...+\dfrac{1}{2^8}\)
\(\Rightarrow2A-A=\left(2+1+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)\)
\(\Rightarrow A=2-\dfrac{1}{2^9}\)
\(\Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)=6-\dfrac{3}{2^9}\)
Vậy...
10 tháng 5 2017
=> S = 3 \(\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\right)\)
S = 3 \(\left(2+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+...+\dfrac{1}{2^8}-\dfrac{1}{2^9}\right)\)
= 3\(\left(2-\dfrac{1}{2^9}\right)\)= 6 \(-\dfrac{3}{2^9}\)
TICK CHO MK NHA!
NT
2
PT
28 tháng 3 2017
Giải:
\(S=3.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
\(2S=3.\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)\)
\(2S-S=3.\left[\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\right]\)
\(S=3.\left(2-\dfrac{1}{2^9}\right)\)
\(S=3.\dfrac{1023}{512}\)
\(S=\dfrac{3069}{512}\)
28 tháng 3 2017
2.S= \(6\) + \(3\) + \(\dfrac{3}{2}\) + ..... + \(\dfrac{3}{2^8}\)
2.S - S= ( \(6\) + \(3\) + \(\dfrac{3}{2}\) + ..... + \(\dfrac{3}{2^8}\)) - (\(3\) + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ..... + \(\dfrac{3}{2^9}\))
S= 6 - \(\dfrac{3}{2^9}\)
S= \(\dfrac{6.512}{512}\) - \(\dfrac{3}{512}\) = \(\dfrac{3069}{512}\)
Bn tự rút gọn nha, mk hơi nhát
4 tháng 1 2024
S = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^99 + 1/3^100
3S = 1 +1/3 +1/3^2 +1/3^3 + ... + 1/3^98 +1/3^99
3S - S = ( 1 + 1/3 + 1/3^2 +1/^3 + ... + 1/3^98 +1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + 1/3^4 +... + 1/3^99 + 1/3^100 )
2S = 1 - 1/3^100
S = (1 - 1/3^100). 1/2
NA
0
17 tháng 7 2021
Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
16 tháng 4 2023
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 = \(\dfrac{2^9-3}{2^8}\)
15 tháng 7 2023
Tổng quát:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow S=\dfrac{10}{11}\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 7 2023
Ta có công thức tổng quát như sau:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)
\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)
Áp dụng vào tổng S ta có:
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)
\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)
\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)
Ta có:
S = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ........+ \(\dfrac{3}{2^9}\)
2S = 6 + 3 + \(\dfrac{3}{2^2}\) + ....... + \(\dfrac{3}{2^8}\)
Mà S = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ........+ \(\dfrac{3}{2^9}\)
=> 2S - S = 6 - \(\dfrac{3}{2^9}\)
\(S=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
Đặt : \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ \Rightarrow2.A=2+1+....+\dfrac{1}{2^8}\\ \Rightarrow A=2-\dfrac{1}{2^9}\\ \Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)\)