y=1/2^2+1/3^2.........+1/2013^2 < 1/1.2 + 1/2.3 +.....+ 1/2012.2013
ta có 1/k-1/k+1=1/k(k+1)
suy ra y<1-1/2013
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\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
Tính A = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2013}\left(1+2+...+2013\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)\)=?
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
CHỨNG TỎ : A<-1/2
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
Và B =\(\frac{1}{2}\)
So sánh A và B
Câu 1 Tính giá trị biểu thức:a) A 2+5+8+11+...+2012b) Bleft(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right).....left(1-frac{1}{2011}right).left(1-frac{1}{2012}right)Câu 2a) Tìm x,y nguyên biết : 2x.(3y-2) + (3y-2) -55b) CMR:frac{1}{4^2}+frac{1}{6^2}+frac{1}{8^2}+...+frac{1}{left(2nright)^2} frac{1}{4}
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Câu 1 Tính giá trị biểu thức:
a) A= \(2+5+8+11+...+2012\)
b) B=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2011}\right).\left(1-\frac{1}{2012}\right)\)
Câu 2
a) Tìm x,y nguyên biết : 2x.(3y-2) + (3y-2) = -55
b) CMR:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(a,-\frac{2}{3}.\left(x-\frac{1}{4}\right)=\frac{1}{3}.\left(2x+1\right)\)
\(b,\frac{1}{5}.2^x+\frac{1}{3}.2^{x+1}=\frac{1}{5}.2^7+\frac{1}{3}:2^8\)
\(c,\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
Cho \(A=\left(\frac{1}{^{2^2}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)va\)\(B=-\frac{1}{2}.\)Hay so sanh A va B