S=1+2+2^2-2^3-...+2^10
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Những câu hỏi liên quan
N
0
15 tháng 10 2023
1:
\(S=-\left(1-\dfrac{1}{10}+\dfrac{1}{10^2}-...-\dfrac{1}{10^{n-1}}\right)\)
\(=-\left[\left(-\dfrac{1}{10}\right)^0+\left(-\dfrac{1}{10}\right)^1+...+\left(-\dfrac{1}{10}\right)^{n-1}\right]\)
\(u_1=\left(-\dfrac{1}{10}\right)^0;q=-\dfrac{1}{10}\)
\(\left(-\dfrac{1}{10}\right)^0+\left(-\dfrac{1}{10}\right)^1+...+\left(-\dfrac{1}{10}\right)^{n-1}\)
\(=\dfrac{\left(-\dfrac{1}{10}\right)^0\left(1-\left(-\dfrac{1}{10}\right)^{n-1}\right)}{-\dfrac{1}{10}-1}\)
\(=\dfrac{1-\left(-\dfrac{1}{10}\right)^{n-1}}{-\dfrac{11}{10}}\)
=>\(S=\dfrac{1-\left(-\dfrac{1}{10}\right)^{n-1}}{\dfrac{11}{10}}\)
2:
\(S=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^{n-1}\)
\(u_1=1;q=\dfrac{1}{3}\)
\(S_{n-1}=\dfrac{1\cdot\left(1-\left(\dfrac{1}{3}\right)^{n-1}\right)}{1-\dfrac{1}{3}}\)
\(=\dfrac{3}{2}\left(1-\left(\dfrac{1}{3}\right)^{n-1}\right)\)
Y
15 tháng 10 2023
\(1,\) Ta có \(\left\{{}\begin{matrix}q=\dfrac{u_2}{u_1}=\dfrac{1}{10}:\left(-1\right)=-\dfrac{1}{10}\\u_1=-1\end{matrix}\right.\)
Vậy \(S=-1+\dfrac{1}{10}-\dfrac{1}{10^2}+...+\dfrac{\left(-1\right)^n}{10^{n-1}}=\dfrac{-1}{1-\left(-\dfrac{1}{10}\right)}=-\dfrac{10}{11}\)
\(2,\) Ta có \(\left\{{}\begin{matrix}q=\dfrac{u_2}{u_1}=\dfrac{1}{3}\\u_1=1\end{matrix}\right.\)
Vậy \(S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}\)
HA
1
TG
10 tháng 4 2022
S = 3 / 10^2 + 3 / 11^2 + 3 / 12^2 +.... 3 / 101^2
=>S<3/9x10+3/10x11+3/11x12+...+3/100x101
=>S<3/9-1/10+1/10-1/11+1/11-1/2+...+1/100-1/101
=>S<1/3-1/101<1/3
Vậy S<1/3
CR
2
15 tháng 1 2017
S=2+2^1+2^2+...+2^100
2S=2.(2+2^1+2^2+...+2^100)
2S=2^2+2^3+...+2^101
2S-S=(2^2+2^3+...+2^101)-(2+2^1+...+2^100)
S=2^101-2
Sx10=(2^101-2)x10
NT
20 tháng 8 2017
S=2+2^1+2^2+2^3+2^4+2^5+.....+2^99+2^100
\(\Rightarrow\)2S = 4 + 22 + 23 + 24 + 25 + 26 + ...... + 2100 + 2101
\(\Rightarrow\)2S - S = ( 4 + 22 + 23 + 24 + 25 + 26 + ...... + 2100 + 2101 ) - ( 2 + 21 + 22+ 23 + 24 + 25 +.....+ 299 + 2100 )
\(\Rightarrow\)S = ( 4 + 2101 ) - ( 2 + 21 )
\(\Rightarrow\)S x 10 = 2101 x 10
Nếu mình đúng thì các bạn k mình nhé
LT
3
6 tháng 4 2019
Ta có: \(\frac{1}{2^2}>\frac{1}{2.3}\)
\(\frac{1}{3^2}>\frac{1}{3.4}\)
.... .... ..........
\(\frac{1}{10^2}>\frac{1}{10.11}\)
\(\Rightarrow S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\left(đpcm\right)\)
Cảm ơn
KN
7 tháng 4 2019
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(\Leftrightarrow S>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
\(\Leftrightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow S>\frac{1}{2}-\frac{1}{11}\)
\(\Leftrightarrow S>\frac{11}{22}-\frac{2}{22}\)
\(\Leftrightarrow S>\frac{9}{22}\left(đpcm\right)\)
YP
2
28 tháng 3 2017
\(A=\dfrac{2^{10}+1}{2^{10}-1},B=\dfrac{2^{10}-1}{2^{10}-3}\)
Dễ thui
\(A=\dfrac{2^{10}-1+2}{2^{10}-1}=1+\dfrac{2}{2^{10}-1}\)
\(B=\dfrac{2^{10}-1}{2^{10}-3}=\dfrac{2^{10}-3+2}{2^{10}-3}=1+\dfrac{2}{2^{10}-3}\)
Vì \(2^{10}-1>2^{10}-3\) nên \(\dfrac{2}{2^{10}-1}< \dfrac{2}{2^{10}-3}\)
Suy ra A<B
31 tháng 1 2022
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
DK
18 tháng 9 2022
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)