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5 tháng 4 2015

\(S=\frac{2^2}{\left(2-1\right)\left(2+1\right)}+\frac{3^2}{\left(3-1\right)\left(3+1\right)}+...+\frac{2008^2}{\left(2008-1\right)\left(2008+1\right)}\)

\(S=\frac{2^2}{2^2-1}+\frac{3^2}{3^2-1}+...+\frac{2008^2}{2008^2-1}=\frac{2^2-1+1}{2^2-1}+\frac{3^2-1+1}{3^2-1}+...+\frac{2008^2-1+1}{2008^2-1}\)

\(S=1+\frac{1}{1.3}+1+\frac{1}{2.4}+...+1+\frac{1}{2007.2009}=\left(1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{2007.2009}\right)\)Tính \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{2007.2009}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{2007.2009}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\right)=\frac{1}{2}.\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\right)\)

\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{2008}-\frac{1}{2009}\right)=...\)

Vậy \(S=2007+A=...\)

 

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 1 2018

A=2^2/1.3+3^2/2.4+4^2/3.5+....+99^2/98.100

A=2^2/(2-1)(2+1)+3^2/(3-1)(3+1)+4^2/(4-1)(4+1)+...+99^2/(99-1)(99+1)

A=2^2/2^2-1+3^2/3^2-1+...+99^2/99^2-1

A=2^2-1+1/2^2-1+3^2-1+1/3^2-1+...+99^2-1+1/99^2-1

A=1+1/1.3+1+1/2.4+1+1/3.5+...+1+1/98.100

A=(1+1+1+....+1)+(1/1.3+1/2.4+...+1/98.100) (1)

Ta có:

Đặt B=(1+1+1+...+1)=98[vì (99-2):1+1=98 số] (2)

Đặt C=1/1.3+1/2.4+1/3.5+...+1/98.100

=>C=1/2.(1-1/3)+1/2.(1/2-1/4)+1/2.(1/3-1/5)+...+1/2.(1/98-1/100)

=>C=1/2.(1-1/3+1/2-1/4+1/3-1/5+...+1/97-1/99+1/98-1/100)

=>C=1/2.(1+1/2-1/99-1/100)

=>C=1/2.(3/2-1/99.100) (3)

Thay (2),(3) vào(1), được:

A=98+1/2.(3/2-1/99.100)

16 tháng 2 2019

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11 tháng 11 2015

Công thức tống quát:

\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}=1+\frac{1}{n^2-1}=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}\)

Theo đó, ta có:

\(1+\frac{1}{1.3}=1+\frac{1}{\left(2-1\right)\left(2+1\right)}=\frac{2^2}{2^2-1}\)

\(1+\frac{1}{2.4}=1+\frac{1}{\left(3-1\right)\left(3+1\right)}=\frac{3^2}{3^2-1}\)

\(1+\frac{1}{3.5}=\frac{1}{\left(4-1\right)\left(4+1\right)}=\frac{4^2}{4^2-1}\)

\(....................\)

\(1+\frac{1}{2015.2017}=1+\frac{1}{\left(2016-1\right)\left(2016+1\right)}=\frac{2016^2}{2016^2-1}\)

Nhân lần lượt các đẳng thức trên, ta được:

\(S=\frac{\left(2.3.4....2016\right)^2}{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)...\left(2016^2-1\right)}=\frac{2^2.3^2.4^2...2016^2}{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(2015.2017\right)}=\frac{2^2.3^2.4^2...2016^2}{1.2.3^2.4^2.5^2...2014^2.2015^2.2016.2017}=\frac{2.2016}{2017}\)

 

 

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

24 tháng 3 2019

\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)

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

\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)

Tự làm tiếp :)) 

tớ nhầm đoạn này tí :((

\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)

\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)

\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính