S=1-2+3-4+5-6+........+2013-2014+2015
Lam tung buoc rui minh tick cho ha may ban . Làm cách nào mà mấy bạn ra số 1007 vậy , chỉ mình đi
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Ax1007x1008=A1= 1007x(1+1/3+...+1/2013)
Bx1007x1008=B1=1008x(1/2+1/4+...+1/2014)
A1-B1=1007x(1-1/2+1/3-1/4+..+1/2013-2/1014) - ( 1/2+1/4+..1/2014)
=1007x(1/2+1/3x4+..1/1007x1008)- (1/2+1/4+..1/2014)
Xet' (1/2+1/4+..1/2014) < (1/2 + 1/2 + .... 1/2) (co' 1007 so' ) = 1007/2
xet' 1007x(1/2 +1/3x4 +... 1/1007x1008 ) > 1007/2
=> A> B
Ta có :
\(S=\left(1+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}=P\)
\(\Rightarrow\left(s-p\right)^{2013}=0^{2013}=0\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^8+3^9\right)=\)
\(=4+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)=\)
\(=4\left(1+3^2+3^4+...+3^8\right)⋮4\)
S = 1 - 2 + 3 - 4 + ..... + 2013 - 2014 + 2015
S = (1 - 2) + (3 - 4) + .... + (2013 - 2014) + 2015
S = -1 + (-1) +.... + (-1) + 2015
S = -1 x 1007 + 2015
S = -1007 + 2015 = 1008