Lời giải:
\(A=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2011}{1.2.3...2012}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{2012-1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2011}-\frac{1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2...2012}< 1\)
Ta có đpcm.
cmr:\(\dfrac{1}{1.2}+\dfrac{2}{1.2.3}+\dfrac{3}{1.2.3.4}+....+\dfrac{2011}{1.2...2012}< 1\)
cmr:\(a=1.2.3...2003.2004.\left(1+\dfrac{1}{2}+\dfrac{1}{2}+...+\dfrac{1}{2003}+\dfrac{1}{2004}\right)\)chia hết cho 2005
CMR biểu thức \(N=\sqrt{1+2011^2+\dfrac{2011^2}{2012^2}}+\dfrac{2011}{2012}\) có giá trị là 1 số tự nhiên
Chứng minh: \(A=1.2.3.....2017.2018\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)⋮2019\)
cmr:a=\(1.2.3...2003.2004.\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2003}+\dfrac{1}{2004}\right)⋮2005\)
Tìm số tự nhiên n sao cho:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
cho \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\). hãy tính giá trị biểu thức sau: \(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
cmr:
Tìm x,y,z thỏa mãn
\(\dfrac{\sqrt{x-2010}-1}{x-2010}+\dfrac{\sqrt{y-2011}-1}{y-2011}+\dfrac{\sqrt{z-2012}-1}{z-2012}=\dfrac{3}{4}\)