chứng minh rằng \(1\cdot3\cdot5\cdot....\cdot2013\cdot2015+2\cdot4\cdot6\cdot....\cdot2014\cdot2016⋮9911\)
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
Tính B=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{2015^2}{2014\cdot2016}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)
\(B=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)
\(B=2015.\frac{1}{1008}\)
\(B=\frac{2015}{1008}\)
Cho \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{2014\cdot2015\cdot2016}\).
Chứng minh \(A\le\dfrac{1}{4}\).
A=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2014\cdot2015\cdot2016}=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2014\cdot2015}-\dfrac{1}{2015\cdot2016}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2015}\cdot\dfrac{1}{2016}\right)=\dfrac{1}{4}-\dfrac{1}{2\cdot2015\cdot2016}< \dfrac{1}{4}\)
Vậy A<\(\dfrac{1}{4}\)
---bé hơn hoặc bằng tức là chỉ cần xảy ra 1 khả năng cũng thõa mãn nhé---
Tìm 3 chữ số tận cùng của :
\(A=1\cdot3\cdot5\cdot7\cdot....\cdot2013\cdot2015\)
vì có số 2015 có c/s tận cùng là 5 suy ra tích trên có c/s t /c là 5
mik nhanh nhất đó
k nhanha
A = \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot.....\cdot\left(1+\frac{1}{2011\cdot2013}\right)\)
Tính
\(A=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+2014\cdot2015\cdot2016\)
\(A=1.2.3+2.3.4+3.4.5+....+2014.2015.2016\)
\(4A=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+....+2014.2015.2016.\left(2017-2013\right)\)\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+2014.2015.2016.2017-2013.2014.2015.2016\)\(4A=2014.2015.2016.2017\)
\(A=\dfrac{2014.2015.2016.2017}{4}=4215446423280\)
Cho \(A=\frac{1\cdot3\cdot5\cdot...\cdot995\cdot997}{4\cdot6\cdot8\cdot...\cdot998\cdot1000};B=\frac{2\cdot4\cdot6\cdot...\cdot996\cdot998}{5\cdot7\cdot9\cdot...\cdot999\cdot1001}\)So sánh A và B
\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{2011\cdot2012\cdot2013\cdot2014}\)
1/1.2.3.4+1/2.3.4.5+....+1/2011.2012.2013.2014
Gọi dãy số trên là A , ta có :
3A=3/1.2.3.4+3/2.3.4.5+....+3 / 2011.2012.2013.2014
3A=1/1.2.3-1/2.3.4+1/2.3.4-......-1/2011.2012.2013+1 / 2011.2012.2013 - 1/2012.2013.2014
3A=1/1.2.3-1/2012.2013.2014
phần còn lại bn tính tiếp đi
Tìm 3 chữ số tận cùng ( BA CHỮ SỐ TẬN CÙNG) của
\(A=1\cdot3\cdot5\cdot7\cdot....\cdot2013\cdot2015\)