\(M=\frac{1.3.5...2011.2013}{1008.1009.1010...2013.2014}\)

\(M=\frac{1.2.3.4.5.6...2011.2012.2013.2014}{\left(2.4.6...2014\right).1008.1009.1010....2013.2014}\)

\(M=\frac{1.2.3.4.5.6...2011.2012.2013.2014}{2^{1007}.\left(1.2.3...1007\right).1008.1009.1010...2013.2014}\)

\(M=\frac{1}{2^{1007}}\)

\(2015=5.13.31\)

Ta có: \(1.2.....1007=1.2...5....13.....31...1007\text{ chia hết cho }5.13.31=2015\)

\(1008.1009.....2004=1008....\left(1010\right)....\left(1014\right)...\left(1023\right)....2004\)

\(=1008....\left(5.202\right)....\left(13.78\right)....\left(31.33\right)...2004\text{ chia hết cho }5.13.33=2015\)

Do đó tổng 2 số trên chia hết cho 2015.

c,

Tổng P có số số hạng là

   (113-2):3+1=38(số)

Có số cặp là

    38:2=19(cặp)

Ta có: P=2-5+8-11+14-17+...+110-113

= (2-5)+(8-11)+(14-17)+...+(110-113)

= (-3)+(-3)+(-3)+...+(-3)

= (-3).19

= -57

https://hoc24.vn/hoi-dap/question/598367.html

\(B=\frac{1.3.5....2011.2013}{1008.1009....2013.2014}\)

\(\Rightarrow B=\frac{1.3.5...1007.1009...2013}{1008.1009...2013.2014}\)

\(\Rightarrow B=\frac{1.3.5...1006}{1008.1010...2014}\)

=1/2^1007  nha

C = \(\frac{1.3.5.......2011.2013}{1008.1009......2013.2014}\)

C =  \(\frac{1.2.3.4.5......2013.2014}{\left(2.4.6....2012.2014\right).\left(1008.1009.....2014\right)}\)

C = \(\frac{1.2.3.......2013.2014}{2^{1007}.\left(1.2.3.4.5.....1007\right).1008.1009....2014}\)

C = \(\frac{1.2.3.4.5.......2014}{2^{1007}.1.2.3.4.5.......2014}\)

C = \(\frac{1}{2^{1007}}\)

Ta có : \(A\text{=}\dfrac{2013.2014-1}{2013.2014}\text{=}\dfrac{2013.2014}{2013.2014}-\dfrac{1}{2013.2014}\text{=}1-\dfrac{1}{2013.2014}\)

\(B\text{=}\dfrac{2014.2015-1}{2014.2015}\text{=}\dfrac{2014.2015}{2014.2015}-\dfrac{1}{2014.2015}\text{=}1-\dfrac{1}{2014.2015}\)

\(Ta\) có : \(\dfrac{1}{2013.2014}>\dfrac{1}{2014.2015}\)

\(\Rightarrow A< B\)

\(A=1-\dfrac{1}{2013.2014}\)

\(B=1-\dfrac{1}{2014.2015}\)

Ta có: \(2014.2015>2013.2014\)

\(\Rightarrow A>B\)