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A=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(22+1)(24+1)(28+1)(216+1)
=[(22-1)(22+1)](24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
Ai tích mk tích lại cho nha !!!
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
\(A=3+3^2+...+3^{50}\)
\(\Rightarrow3A=3^2+3^3+...+3^{50}+3^{51}\)
\(\Rightarrow3A-A=3^{51}-3\)
\(\Rightarrow2A=3^{51}-3\)
\(\Rightarrow A=\frac{3^{51}-3}{2}\)
\(B=2-2^2+2^3-2^4+...+2^{2019}-2^{2020}\)
\(2B=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(B+2B=2-2^{2021}\)
\(3B=2-2^{2021}\)
\(B=\frac{2-2^{2021}}{3}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(C=1-\frac{1}{2009}\)
\(C=\frac{2008}{2009}\)
\(D=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(D=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(D=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(D=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
Ta có : \(A=3+3^2+3^3+.....+3^{2016}\)
\(\Rightarrow3A=3^2+3^3+3^4+......+3^{2017}\)
\(\Rightarrow3A-A=3^{2017}-3\)
\(\Rightarrow2A=3^{2017}-3\)
\(\Rightarrow A=\frac{3^{2017}-3}{2}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{1024}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{512}\)
\(\Rightarrow2B-B=1-\frac{1}{1024}\)
\(\Rightarrow B=\frac{1023}{1024}\)
đó giúp mk đi mà
à, mk quên chưa nói là ai giúp mk sẽ được luôn 2SP đó
giúp mk nha
cảm ơn nhiều!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a: \(S=\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(=4\left(1+3^2+3^4+...+3^8\right)⋮4\)
b: \(S=\left(1+2\right)+2^2\left(1+2\right)+...+2^8\left(1+2\right)\)
\(=3\left(1+2^2+...+2^8\right)⋮3\)
a: Ta có: \(A=\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2020}}\)
=>\(2A=1+\frac12+\cdots+\frac{1}{2^{2019}}\)
=>\(2A-A=1+\frac12+\cdots+\frac{1}{2^{2019}}-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{2020}}\)
=>\(A=1-\frac{1}{2^{2020}}=\frac{2^{2020}-1}{2^{2020}}\)
b: \(B=1+\frac12+\frac14+\ldots+\frac{1}{2048}\)
=>\(B=1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{11}}\)
=>\(2B=2+1+\frac12+\cdots+\frac{1}{2^{10}}\)
=>\(2B-B=2+1+\frac12+\cdots+\frac{1}{2^{10}}-1-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{11}}\)
=>\(B=2-\frac{1}{2^{11}}=\frac{2^{12}-1}{2^{11}}\)
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