Chứng minh rằng với mọi số tư nhiên n khác 0 ta đều có:1/2.5+1/5.8+1/8.11+...+1/(3n-1)(3n+2)=n/6n+4
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HA
28 tháng 1 2016
Đặt A=1/2.5+1/5.8+...+1/(3n-1).(3n+2)
=>3A=3/2.5+3/5.8+...+3/(3n-1).(3n+2)
=>3A=1/2-1/5+1/5-1/8+...+1/3n-1-1/3n+2
=>3A=1/2-1/3n+2
=>3A=(3n+2-2)/[2.(3n+2)]
=>3A=3n/6n+4
=>A=3n/6n+4/3
=>A=n/6n+4
14 tháng 4 2016
Đặt \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=>3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\)
=> \(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{3n-1}-\frac{1}{3n+2}\)
=>\(3A=\frac{1}{2}-\frac{1}{3n+2}\)
=> \(3A=\frac{\left(3n+2\right):2}{3n+2}-\frac{1}{3n+2}\)
=> \(3A=\frac{1,5.n}{3n+2}\)
=>\(A=\frac{1,5.n}{3n+2}.\frac{1}{3}=>A=\frac{1,5.n}{\left(3n+2\right).3}=\frac{1,5.n}{9n+6}\)
\(Hay\) \(A=\frac{1,5n:1,5}{\left(9n+6\right):1,5}=\frac{n}{9n:1,5+6:1,5}=\frac{n}{6n + 4} \left(đpcm\right)\)
IL
2 tháng 7 2018
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)\)
\(=\dfrac{1}{3}.\dfrac{3n}{6n+4}\)
\(=\dfrac{n}{6n+4}\) ( đpcm )
Vậy...
10 tháng 10 2017
Đặt :
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+............+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+........+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{3n+2}\)
FT
10 tháng 10 2017
@Akai Haruma em không hiểu tại sao bài kia chị lại tick cho bạn đó ạ,đề nói chứng minh,mak bạn đó đã làm hết đâu:
\(VT=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{3n-1}+\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{6}-\dfrac{1}{9n+6}\)
\(VT=\dfrac{9n+6}{54n+36}-\dfrac{6}{54n+36}\)
\(VT=\dfrac{9n+6-6}{54n+36}=\dfrac{9n}{54n+36}=\dfrac{9n}{9\left(6n+4\right)}=\dfrac{n}{6n+4}=VP\left(đpcm\right)\)
22 tháng 3 2017
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự
Đặt A=1/2.5+1/5.8+...+1/(3n-1)(3n+2)
3A=3/2.5+3/5.8+....+3/(3n-1)(3n+2)
3A=1/2-1/5+1/5-1/8+....+1/3n-1-1/3n+2
3A=1/2-1/3n+2
3A=3n/6n+4
A=(3n/6n+4) /3
A=n/6n+4(đpcm)