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\(B=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+4\right)+...+2^{96}\left(1+2+2^2\right)\)
\(=7\left(1+2^4+2^8+...+2^{96}\right)⋮7\)
đặt biểu thức trên là A
ta có :
A = 1 + 3 +32 + 33 + 34 + ... + 310
3A = 3 + 32 + 33 + 34 + 35 + ... + 311
3A - A = ( 3 + 32 + 33 + 34 + 35 + ... + 311 ) - ( 1 + 3 +32 + 33 + 34 + ... + 310 )
2A = 311 - 1
A = ( 311 - 1 ) : 2 => điều phải chứng tỏ
a) Ta thấy: 1/2^2<1/1.2
1/3^2<1/2.3
1/4^2<1/3.4
…………...
1/100^2<1/99.100
=>A<1/1.2+1/2.3+1/3.4+…+1/99.100=99/100
Mà 99/100<1 => 1/22 + 1/32 + 1/42 + ... + 1/1002<1
b)Ta thấy : 1/101+1/102+1/103+…+1/150>1/150+1/150+1/150+…+1/150(50 số hạng)
=>A>50/150>1/3 (1)
Ta thấy : 1/101+1/102+1/103+…+1/150<1/100+1/100+1/100+…+1/100(50 số hạng)
=>A<1/2 (2)
Từ (1) và (2) =>1/3<A<1/2
c) Ta thấy : 1/11 + 1/12 + 1/13 + ... + 1/20>1/20+1/20+1/20+…+1/20(10 số hạng)
=>1/11 + 1/12 + 1/13 + ... + 1/20>1/2
B = 1 + 4 + 42 +...+ 4200 + 4201
=> 4B = 4 + 42 +43 +...+ 4201 + 4202
=> 4B-B = 4202 - 1
3B = 4202 -1
\(\Rightarrow B=\frac{4^{202}-1}{3}\)
4B = 4 + 4^2 + 4^3 + ... + 4^202
4B - B = ( 4 + 4^2 + 4^3 + ... + 4^202 ) - ( 1 + 4 + 4^2 + ... + 4^201 )
3B = 4^202 - 1
B = \(\frac{4^{202}-1}{3}\)
\(2.\left(2x-\frac{4}{3}\right)^2+\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)^2=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)=\sqrt{\frac{1}{4}}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}+\frac{4}{3}\)
\(\Rightarrow2x=\frac{11}{6}\)
\(\Rightarrow x=\frac{11}{6}\div2\)
\(\Rightarrow x=\frac{11}{6}\times\frac{1}{2}\)
\(\Rightarrow x=\frac{11}{12}\)
Đặt \(A=1+2+2^2+2^3+...+2^{20}\)
\(2A=2+2^2+2^3+2^4+...+2^{21}\)
\(2A-A=\left(2+2^2+2^3+...+2^{21}\right)-\left(1+2+2^2+...+2^{20}\right)\)
\(A=2^{21}-1\)
Ta đặt
A= 1+2^1+2^2+2^3+....2^20
2A= 21+22+23+....+221
=>2A-A=(2^1+2^2+2^3+...+2^21)-(1+2^2+2^3+...)
1A=2^21-1
Vậy A=2^21-1
B = (1 - \(\dfrac{1}{2^2}\)).(1 - \(\dfrac{1}{3^2}\)).(1 - \(\dfrac{1}{4^2}\))...(1 - \(\dfrac{1}{201^2}\))
B = \(\dfrac{2^2-1}{2^2}\).\(\dfrac{3^2-1}{3^2}\).\(\dfrac{4^2-1}{4^2}\)...\(\dfrac{201^2-1}{201^2}\)
B = \(\dfrac{4-1}{2^2}\).\(\dfrac{9-1}{3^2}\).\(\dfrac{16-1}{4^2}\)...\(\dfrac{40401-1}{201^2}\)
B = \(\dfrac{3}{2^2}\).\(\dfrac{8}{3^2}\).\(\dfrac{15}{4^2}\)....\(\dfrac{40400}{201^2}\)
B = \(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\).\(\dfrac{3.5}{4.4}\).\(\dfrac{4.6}{5.5}\)...\(\dfrac{200.202}{201.201}\)
B = \(\dfrac{1}{2}\).\(\dfrac{202}{201}\)
B = \(\dfrac{101}{201}\)