K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

Đặt A = \(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2010.2012+2}{2011^2}+\frac{2015.2017+2}{2016^2}\)

\(=\frac{\left(2-1\right)\left(2+1\right)+2}{2^2}+\frac{\left(3-1\right)\left(3+1\right)}{3^2}+...+\frac{\left(2016-1\right)\left(2016+1\right)+2}{2016^2}\)

\(=\frac{2^2-1+2}{2^2}+\frac{3^2-1+2}{3^2}+....+\frac{2016^2-1+2}{2016^2}\)

\(=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+...+\frac{2016^2+1}{2016^2}\)

\(=\left(1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\right)\)(2015 hạng tử 1)

\(=2015+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2016.2016}\right)\)

 \(< 2015+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)

\(=2015+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)=2015+\left(1-\frac{1}{2016}\right)\)

= 2015 + 1 + 1/2016

= 2016 + 1/2016 < 2017 

=> A < 2017 (ĐPCM)

18 tháng 3 2016

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

23 tháng 10 2016

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

17 tháng 6 2016

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.,,\frac{50^2}{49.51}\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.,,\frac{50.50}{49.51}\)

=\(\frac{\left(2.3.4...50\right).\left(2.3.4...50\right)}{\left(1.2.3....49\right).\left(3.4.5....51\right)}\)

=\(\frac{50.2}{1.51}\)

=\(\frac{100}{51}\)

17 tháng 6 2016

\(=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4...50}{3.4.5...51}=50.\frac{2}{51}=\frac{100}{51}\)

9 tháng 3 2021

\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

3 tháng 5 2018

mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha

25 tháng 5 2022

\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

25 tháng 5 2022

undefined

\(=\dfrac{1}{2}\cdot\dfrac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2016^2-1+1}{\left(2016-1\right)\left(2016+1\right)}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2016}{2015}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2016}{2017}\)

\(=\dfrac{1}{2}\cdot2016\cdot\dfrac{2}{2017}=\dfrac{2016}{2017}\)