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a) Đặt A = 1 + 2 + 22 + ... + 22008 (1)
=> 2A = 2 + 22 + 23 + ... + 22009 (2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = (2 + 22 + 23 + ... + 22009) - (1 + 2 + 22 + ... + 22008)
A = 22009 - 1
Khi đó B = \(\frac{2^{2009}-1}{1-2^{2009}}=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=-1\)
b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}\)
=> A - 1 = \(\frac{20^{10}+1-20^{10}+1}{20^{10}}=\frac{2}{20^{10}}\)
Lại có B = \(\frac{20^{10}-1}{20^{10}-3}\)
=> B - 1 = \(\frac{20^{10}-1-20^{10}+3}{20^{10}-3}=\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{2^{10}}< \frac{2}{2^{10}-3}\)
=> A - 1 < B - 1
=> A < B
a) \(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Đặt \(Q=1+2+2^2+...+2^{2008}\)
\(2Q=2+2^2+2^3+...+2^{2009}\)
\(2Q-Q=2+2^2+2^3+...+2^{2009}-1-2-2^2-...-2^{2008}\)
\(\Rightarrow Q=2^{2009}-1\)
Ta thấy \(Q\) là số đối của \(2^{2009}-1\)
\(\Rightarrow B=-1\)
Vậy \(B=-1\).
b) Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
Ta lại có: \(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\) nên \(1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
a/A=1+2+4+8+...+1024
2A=2+4+8+16+....+2048
2A-A=(2+4+8+16+....+2048)-(1+2+4+8+...+1024)
A=2048-1
A=2047
VẬY A=2047
b/B=1+5+25+125+....+15625
5B=5+25+125+625+....+78125
5B-B=(5+25+125+625+....+78125)-(1+5+25+125+....+15625)
4B=78125-1
4B=78124
B=78124:4
B=19531
VẬY B =19531
C=1/1.2+1/2.3+1/3.4+...+1/2015.2016
C=1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016
=1-1/2016
=2015/2016
VẬY C=2015/2016
D/=10/1.3+10/3.5+10/5.7+....+10/2013.2015
=5(2/1.3+2/3.5+2/5.7+...+2/2013.2015)
=5(1-1/3+1/3-1/5+1/5-1/7+..+1/2013-1/2015)
=5(1-1/2015)
=5.2014/2015
=2014/403
VẬY D=2014/403
a, A = 1 + 2 + 4 + 8 +...+ 1024
\(A=1+2+2^2+2^3+....+2^{10}\)
\(2A=2+2^2+2^3+....+2^{10}+2^{11}\)
\(A=1+2+2^2+2^3+....+2^{10}\)
\(A=2^{11}-1=2047\)
b, B = 1 + 5 + 25 + 125 + ... + 15625
\(B=1+5+5^2+5^3+....+5^6\)
\(3B=5+5^2+5^3+....+5^6+5^7\)
\(B=1+5+5^2+5^3+....+5^6\)
\(2B=5^7-1\Rightarrow B=\frac{5^7-1}{2}=39062\)
d, D = 10 / 1 . 3 + 10 / 3 . 5 + 10 / 5 . 7 + ... + 10 / 2013 . 2015
\(D=\frac{10}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(D=\frac{10}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(D=\frac{10}{2}.\left(1-\frac{1}{2015}\right)=5.\frac{2014}{2015}=\frac{2014}{403}\)
Câu c thì tương tự
Đặt A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
3A=3/2.5+3/5.8+....+3/17.20
3A=1/2-1/5+1/5-1/8+...+1/17-1/20
3A=1/2-1/20
3A=9/20
2)
Giữ nguyên p/s 1/2^2
Ta có:1/3^2<1/2.3
1/4^2<1/3.4
...............
1/n^2<1/(n-1).n
=>1/3^2+1/4^2+...+1/n^2<1/2.3+1/3.4+...+1/(n-1).n
=>1/3^2+1/4^2+.....+1/n^2<1/2-1/3+1/3-1/4+.........+1/n-1-1/n
=>1/2^2+1/3^2+.....+1/n^2<1/2^2+1/2-1/n
=>1/2^2+1/3^2+....+1/n^2<3/4-1/n<3/4
3)
2B=2/3.5+2/5.7+....+2/47.49+2/49.51
2B=1/3-1/5+1/5-1/7+.....+1/47-1/49+1/49-1/51
2B=1/3-1/51
2B=16/51
B=16/51:2
B=8/51
A=1+1/2+1/2^2+...+1/2^2010
2A=2+1+1/2+....+1/2^2009
2A-A=(2+1+1/2+...+1/2^2009)-(1+1/2+1/2^2+....+1/2^2010)
A=2-1/2^2010
Ta có:
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{11}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
a=1/2+1/2^2+.........+1/2^10
2a=1+1/2+1/2^2+........+1/2^9
2a-a=1-2^10
=1023/1024
2A=2*(1/2+1/2^2+....+1/2^10)
2A=1+1/2+1/2^2+....+1/2^9
2A - A =A=1 - 1/2^10