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a/ 2H=2^2011-2^2010-2^2009-...-2
=> 2H-H=2^2011-2^2010-2^2009-...-2-(2^2010-2^2009-2^2008-...-1)
H=2^2011-2^2010-2^2009-...-2-2^2010+2^2009+2^2008+...+1
H=2^2011-2^2010-2^2010-1
H=2^2011-2.2^2010-1
H=2^2011-2^2011-1
H=-1 => 2010^-1=1/2010
b/ M=1 + 1/2(1+2) + 1/3(1+2+3) + 1/4(1+2+3+4) + ... + 1/16(1+2+3+...+16)
M= 1+1/2.(2.3/2) + 1/3.(3.4/2) + 1/4.(4.5/2) + ... + 1/16.(16.17/2)
M= 1 + 3/2 +4/2 + 5/2 + ... + 17/2
Cùng mẫu số rồi Tự tính nhé
có 1 công thức làm bài này nè em : 1+2=3=2.3/2, 1+2+3=6=3.4/2, 1+2+3+4=10=4.5/2 ....
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(=\frac{\frac{17.18}{2}-1}{2}=76\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(\Rightarrow D=\frac{1}{2012}\)
câu hỏi hay......nhưng tui xin nhường cho các bn khác
Hãy tích đúng cho tui nha
THANKS
\(=1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2}+...+8+\frac{17}{2}\)
\(=\left(1+2+...+8\right)+\left(\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\right)=36+\frac{80}{2}=36+40=76\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)
\(B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(B=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)
\(B=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)
\(S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(A=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2\)
\(\Rightarrow2A-2^{2010}+1=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A-2^{2010}+1=A\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow S=2^{2010}-A=2^{2010}-\left(2^{2010}-1\right)=1\)
b/ Ta có công thức \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
Do đó:
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+...+\dfrac{1+2+3+...+16}{16}\)
\(P=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{2.3}+\dfrac{4.5}{2.4}+...+\dfrac{16.17}{2.16}\)
\(P=1+\dfrac{1}{2}\left(3+4+5+...+17\right)\)
\(P=1+\dfrac{1}{2}.\dfrac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)