cho s = 1+2+2mu2+...+2mu2018 tính s
Biết rằng 1mu 2 +2mu2 +3mu2+......+10mu2= 385 tính nhanh di tổng. S=2mu2+4mu2+6mu2+.....+20mu2
S = 22 + 42 + 62 + ... + 202
= (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2
= 22.12 + 22.22 + 22.32 + ... + 22.102
= 22 (12 + 22 + ... + 102 )
= 4 . 385
= 1540
Cho S =1+2+2mu2+2mu3+...+2mu9+2mu10+2mu11.Hay so sanh Svoi 5*2mu10
\(S=\dfrac{2mu2}{1.2}+\dfrac{2mu2}{2.3}+\dfrac{2mu2}{3.4}+...+\dfrac{2mu2}{2022.2023}\)
(mu = mũ)
\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)
\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)
\(S=2^2.\dfrac{2022}{2023}\)
\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)
S=2+2mu2+2mu3+....+2mu100;hay chung to rang S chia het cho 15 va chu so tan cung cua S
tinh S=1/2+1/2mu2+1/2mu3+...+1/2mu20
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)
=> \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> \(S=1-\frac{1}{2^{20}}\)
11/2mu2+1/3mu2+1/4mu2+...+1/9mu2 cmr 2/5<s<8/9
S=1/2mu2 + 1/3mu2 + 1/4mu2 + ... + 1/100mu2 bé hơn 1
Ta thấy :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
......
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow S< 1-\frac{1}{100}\)
Mà \(1-\frac{1}{100}< 1\)nên \(S< 1\)
Ủng hộ mk nha !!! *_*
cho S = 2+2mu 2+ 2mu3+ .......+2mu2018 a) rút gọn S b) so sanhS voi 2mu 2018 giup minh voi
a) 2S = 22 + 23 + 24 + ... + 22019
2S - S = ( 22 - 22 ) + ( 23 - 23 ) + ... + ( 22018 - 22018) + ( 22019 - 2 )
S = 22019 - 2
b) S > 22018
cho 1phan 2mu2+1phan3mu2+1phan4mu2+....+1phan9mu2 cmr2phan5<s<8phan9