Tính tổng S(n) = \(\dfrac{1}{2,5}+\dfrac{1}{5,8}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\).
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VT
1
17 tháng 12 2022
\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{3n+2-2}{2\left(3n+2\right)}=\dfrac{n}{2\left(3n+2\right)}\)
MH
11 tháng 2 2022
\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)
22 tháng 10 2017
Sửa đề
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{6n+4}\right)=\dfrac{1}{3}\left(\dfrac{3n}{6n+4}\right)=\dfrac{3n}{18n+12}=\dfrac{3n}{3\left(6n+4\right)}=\dfrac{n}{6n+4}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 4 2021
\(\Leftrightarrow\dfrac{u_{n+1}}{n+1}=\dfrac{1}{3}.\dfrac{u_n}{n}\)
Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{3}\\v_{n+1}=\dfrac{1}{3}v_n\end{matrix}\right.\)
\(\Rightarrow v_n\) là CSN với công bội \(\dfrac{1}{3}\)
\(\Rightarrow v_n=\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^{n-1}=\left(\dfrac{1}{3}\right)^n\)
\(S=\sum\limits^{10}_{k=1}\left(\dfrac{1}{3}\right)^k=\dfrac{\dfrac{1}{3}\left(1-\dfrac{1}{3^{10}}\right)}{1-\dfrac{1}{3}}=\dfrac{1}{2}\left(1-\dfrac{1}{3^{10}}\right)\)
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:
\(VT=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{3n-1}+\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{6}-\dfrac{1}{9n+6}\)
\(VT=\dfrac{9n+6}{54n+36}-\dfrac{6}{54n+36}\)
\(VT=\dfrac{9n+6-6}{54n+36}=\dfrac{9n}{54n+36}=\dfrac{9n}{9\left(6n+4\right)}=\dfrac{n}{6n+4}=VP\left(đpcm\right)\)
HT
17 tháng 1 2021
Dang này thì cứ chọn số hạng có mũ cao nhất trên tử và mẫu là được. Nó là ngắt vô cùng lớn hay bé gì đấy
\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 12 2020
\(\dfrac{u_{n+1}}{n+1}=3.\dfrac{u_n}{n}\)
Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{3}\\v_{n+1}=3v_n\end{matrix}\right.\)
\(\Rightarrow v_n=\dfrac{1}{3}.3^{n-1}=3^{n-2}\)
\(\Rightarrow S=3^{-1}+3^0+...+3^8=...\)
HT
16 tháng 2 2021
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
Bn viết để sai rồi, mk sửa lại :)
\(S\left(n\right)=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3S\left(n\right)=\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)
\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{3n+2}\)
\(\Leftrightarrow S\left(n\right)=\dfrac{\dfrac{1}{2}-\dfrac{1}{3n+2}}{3}\)