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A=2^1+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)
=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)
=2.3+2^3.3+...+2^2010.3
=(2+2^3+2^2010).3
=> A chia het cho 3
a,\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=B\left(ĐPCM\right)\)
b, \(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1003}\right)\)
\(A=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)
ui ghi lộn, chữ đpcm chuyển xuống dòng cuối cùng nhé :v
I don't now
mik ko biết
sorry
......................
`@`\(A=1+2^1+2^2+2^3+2^4+2^5\)
\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+2^6\)
\(A=2A-A=\left(2+2^2+2^3+2^4+2^5+2^6\right)-\left(1+2^1+2^2+2^3+2^4+2^5\right)\)
\(A=2^6-1\left(đfcm\right)\)
`@`\(B=1+2^1+2^2+...+2^{2005}\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{2006}\)
\(B=2B-B=\left(2+2^2+2^3+...+2^{2006}\right)-\left(1+2^1+2^2+...+2^{2005}\right)\)
\(B=2^{2006}-1\left(đfcm\right)\)