Chứng minh rằng: 1.3+2/2^2+2.4+2/3^2+3.5+2/4^2+...+2010.2012+2/2011^2+2015.2017+2/2016^2<2017
Mình cần gấp lắm, cố xong trong hôm nay nha
tính A=\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{2016^2}{2015.2017}\)
A=4/3+9/8+16/15+..............+4064256/4064255
A=1+1/3+1+1/8+1/15+...............+1/4064255
A=(1+1+...+1)+(1/3+1/8+...+1/406255) (có 2015 số 1)
A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)
A=2015+1/2(1+1/2-1/2016-1/2017)
A=2015,749504
k cho mình nhé mình k lại cho
Chứng minh rằng :
\(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2011.2013+2}{2011^2}< 2013\)
Theo quy luật mà mình nhận thấy thì 20112 phải sửa thành 20122 bạn ạ!
Đặt \(A=\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2011.2013+2}{2012^2}\)
\(\Leftrightarrow A=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+\frac{4^2+1}{4^2}+...+\frac{2012^2+1}{2012^2}\)
\(\Leftrightarrow A=1+\frac{1}{2^2}+1+\frac{1}{3^2}+1+\frac{1}{4^2}+...+1+\frac{1}{2012^2}\)
\(\Leftrightarrow A=\left(1+1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
\(\Leftrightarrow A=2011+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
Có: \(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2012}\)
\(\Rightarrow A=2011+B< 2011+1-\frac{1}{2012}\)
\(\Rightarrow A< 2012-\frac{1}{2012}< 2013\)
Ta có đpcm
Tính giá trị của biểu thức sau bằng cách hợp lí nhất
D = \(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{2011^2}{2010.2012}\)
D=22/1.3+32/2.4+42/3.5+…+20112/2011.2012
Tính hợp lý
D = \(\frac{2^2}{1.3}\)+ \(\frac{3^2}{2.4}\)+ \(\frac{4^2}{3.5}\)+ ......................+ \(\frac{2001^2}{2010.2012}\)
CHỨNG MINH
\(\dfrac{1.3+2}{2^{2^{ }}}\)+\(\dfrac{2.4+2}{3^2}\)+\(\dfrac{3.5+2}{4^2}\)+...+\(\dfrac{2008.2010+2}{2009^2}\)+\(\dfrac{2009.2011+2}{2010^2}\) < 2011
GIÚP TỚ ĐI MÀ :))
Trước hết ta chứng minh (a-1)(a+1) + 1 = a^2 (*)
Thật vậy VT = (a-1)(a+1)+1=(a-1)a + a-1 +1 = a^2-a+a=a^2 =VP
Áp dụng (*) ta có:
\(A=\dfrac{1\cdot3+2}{2^2}+\dfrac{2\cdot4+2}{3^2}+...+\dfrac{2009\cdot2011+2}{2010^2}\\ =\dfrac{2^2+1}{2^2}+\dfrac{3^2+1}{3^2}+...+\dfrac{2010^2+1}{2010^2}=2009+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2010^2}\\ < 2009+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2009\cdot2010}\\ =2009+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2009}-\dfrac{1}{2010}=2010-\dfrac{1}{2010}< 2020< 2011\)
A= 1/2 ( 1 + 1.3) ( 1 + 1/2.4) ( 1 + 3.5 ) ...............( 1 + 1/ 2015.2017)
1.chứng minh rằng : s = 1/4+1/16+1/36+....+1/100<1/2
2.tính :s = 1.3 +2.4+3.5 +4.6+.....+2016.2018
2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017
=1-1/3+1/3-1/5+...+1/2015-1/2017
=1-1/2017
=2016/2017