Tính:1/1.3 + 1/3.5 +1/5.7 +........+1/(2n-1)(2n+1)?
Tính S = 1.3/3.5 + 2.4/5.7 + 3.5/7.9 + ... + ( n-1)( n+1) / (2n-1)(2n+1) + ... + 1002.1004/2005.2007
\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)
\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)
\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)
\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)
\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)
\(\Rightarrow S=125,4372197\)
\(\)
Tính : 1.3+3.5+5.7+...+(2n+1).(2n+2)=?
1.3+3.5+5.7+...+(2n+1).(2n+3)=(2n+1).(2n+2).(2n+3).(2n+4)
Tình B= 1.3/3.5+2.4/5.7+3.5/7.9+....+(n-1)(n+1)/(2n-1)/2n+1 plzzzz
Tính tổng C = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/(2n-1) ( 2n+1) ( n thuộc N)
Tốt hok giúp vs
bn lên ngạng hoặc và xem câu hỏi tương tự nha!
Nhớ k mk đấy nha!
thanks nhìu!
OK..OK..OK
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(2C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
Ta có :
\(\frac{2}{1.3}=1-\frac{1}{3}\)
\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)
...............................
\(\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(\Rightarrow2C=1-\frac{1}{2n+1}=\frac{2n}{2n+1}\)
\(\Rightarrow C=\frac{n}{2n+1}\)
a)Tính: 1/1.3 + 1/3.5 + 1/5.7 +...+1/19.21
b) chúng minh: A= 1/1.3 + 1/3.5 +...+ 1/(2n-1)(2n+1) < 1/2
a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21
Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21
= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21
= 1-1/21 = 20/21
=> B= 20/21 : 2 => B= 10/21
b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2
=> A= 1/2- 1/2n+1
=> A< 1/2 ( đpcm )
tính tổng:
I=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(\Rightarrow I=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\right)\)
\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\right)\)
\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{2n+3}\right)\)
\(\Rightarrow I=\frac{1}{2}.\frac{2n+2}{2n+3}\)
\(\Rightarrow I=\frac{n+1}{2n+3}\)
\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{1}{1}-\frac{1}{2n+3}\)
\(=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
a,tính \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{19.21}\)
b,CMR A\(=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{(2n-1).(2n+1)}\le\frac{1}{2}\)
tớ làm câu b thôi, câu a nhân 1/2 lên là đc
\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2.n-1}-\frac{1}{2n+1}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)
p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)
CMR : vs mọi n thuc N TA CÓ :
1/1.3 + 1/3.5 + 1/5.7 + ... + 1/(2n+1)(2n+3) = n+1/2n+3
CM: \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\) = \(\dfrac{n+1}{2n+1}\)
Ta có:
VT = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\)+....+\(\dfrac{2}{\left(2n+1\right)\left(2n+3\right)}\))
VT = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+....+ \(\dfrac{1}{2n+1}\) - \(\dfrac{1}{2n+3}\))
VT = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{1}\) - \(\dfrac{1}{2n+3}\) )
VT = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{2n+3}{2n+3}\) - \(\dfrac{1}{2n+3}\))
VT = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2n+2}{2n+3}\)
VT = \(\dfrac{1}{2}\) \(\times\)\(\dfrac{2\times\left(n+1\right)}{2n+3}\)
VT = \(\dfrac{n+1}{2n+3}\) = VP (đpcm)
Chứng minh rằng :
1/1.3 + 1/3.5 + 1/5.7 + ... + 1/(2n-1)(2n+1) < 1/2
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
=\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
=\(\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
=\(\dfrac{1}{2}-\dfrac{1}{4n+2}< \dfrac{1}{2}\)
đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=> 2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
<=> 2A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{2n-2}-\dfrac{1}{2n+1}\)
<=>2A=\(1-\dfrac{1}{2n+1}\)
<=> A=\(\left(1-\dfrac{1}{2n+1}\right)\)\(.\dfrac{1}{2}\)
<=> A=\(\dfrac{1}{2}-\dfrac{1}{2\left(2n+1\right)}\)
=>\(A< \dfrac{1}{2}\) (đpcm)