\(VT=\dfrac{1}{1.5} \dfrac{1}{5.9} ... \dfrac{1}{\left(4n-3\right)\left(4n 1\right)}\) \(VT=\dfrac{1}{4}\left(1-\dfrac{1}{5} \dfrac{1}{5}-\dfrac{1}{9} ... \dfrac{1}{4n-3}-\dfrac{1}{4n 1}\right)\) \(...

MS

Mashiro Shiina

10 tháng 9 2017

anh dz-À mà ko dz đâu .-. xem giúp em đi

NH

Nguyễn Hoàng Linh

27 tháng 11 2017

Tính các tổng sau :

a) \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

b) \(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.11}+...+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)

c) \(\dfrac{7}{1.8}+\dfrac{7}{8.15}+\dfrac{7}{15.22}+...+\dfrac{1}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

5 tháng 6 2022

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

TT

Tạ Thu Phương

3 tháng 12 2017

Tính các tổng sau :

a, \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

b, \(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.11}+........+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)

c,\(\dfrac{7}{1.8}+\dfrac{7}{8.15}+\dfrac{7}{15.22}+....+\dfrac{1}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

7 tháng 6 2022

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)

\(=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

R

Ruby

2 tháng 7 2018

CM: \(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+.....+\dfrac{3}{\left(4n-1\right)\left(4n+3\right)}=\dfrac{5n}{4n+3}\)

#Toán lớp 6

1

J

Jatsumin

2 tháng 7 2018

\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)

Mk thực sự nghĩ đề hình như bị sai hay sao ấy!

Chứng minh rằng với mọi số tự nhiên khác 0 ta đều có : a) \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right).\left(3n+2\right)}=\dfrac{n}{6n+4}\) b) \(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{\left(4n-1\right).\left(4n+3\right)}=\dfrac{5n}{4n+3}\) giúp mk với...

VM

Vũ Minh Hằng

22 tháng 3 2017

Đọc tiếp

#Toán lớp 6

2

N

ngonhuminh

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

VM

Vũ Minh Hằng

25 tháng 3 2017

Cảm ơn bạn

DG

Đừng gọi tôi là Jung Hae Ri

10 tháng 2 2022

Tìm các giới hạn sau:

\(a,\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)

\(b,\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)

3

MH

Minh Hiếu

11 tháng 2 2022

\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)

\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)

Đúng(3)

MH

Minh Hiếu

11 tháng 2 2022

\(a,lim\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)

\(=lim\dfrac{4-\dfrac{3}{n^3}}{\left(3-\dfrac{2}{n^2}\right)\left(\dfrac{1}{n^3}-4\right)}\)

\(=\dfrac{4-0}{\left(3-0\right)\left(0-4\right)}=\dfrac{4}{-12}=-\dfrac{1}{3}\)

SG

Sách Giáo Khoa

10 tháng 4 2017

Cho tổng :

\(S_n=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+....+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)

a) Tính \(S_1,S_2,S_3,S_4\)

b) Dự đoán công thức tính \(S_n\) và chứng minh bằng phương pháp quy nạp

1

BT

Bùi Thị Vân

23 tháng 5 2017

a)
\(S_1=\dfrac{1}{1.5}=\dfrac{1}{5}\)
\(S_2=\dfrac{1}{1.5}+\dfrac{1}{5.9}=\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{5}\right)+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}\right)=\dfrac{1}{4}\left(1-\dfrac{1}{9}\right)=\dfrac{2}{9}\).
\(S_3=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{13}\right)=\dfrac{3}{13}\).
\(S_4=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}\)\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{17}\right)=\dfrac{4}{17}\).
b) Dự đoán công thức : \(S_n=\dfrac{1}{4}\left(1-\dfrac{1}{4n+1}\right)\).
Chứng minh bằng quay nạp:
Với \(n=1\): \(S_1=\dfrac{1}{1.5}=\dfrac{1}{5}\).
Vậy giả thiết quy nạp đúng với n = 1.
Giả sử điều cần chứng minh đúng với \(n=k\).
Nghĩa là: \(S_k=\dfrac{1}{4}\left(1-\dfrac{1}{4k+1}\right)\).
Ta sẽ chứng minh nó đúng với \(n=k+1\): \(S_{k+1}=\dfrac{1}{4}\left(1-\dfrac{1}{4\left(k+1\right)+1}\right)\)
Thật vậy:
\(S_{k+1}=S_k+\dfrac{1}{\left[4\left(k+1\right)-3\right].\left[4\left(k+1\right)+1\right]}\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{4k+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{4\left(k+1\right)-3}-\dfrac{1}{4\left(k+1\right)+1}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{4k+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{4k+1}-\dfrac{1}{4\left(k+1\right)+1}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{4\left(k+1\right)+1}\right)\).
Vậy điều cần chứng minh đúng với mọi n.

DN

đoàn ngọc hân

17 tháng 1 2021

Tính các giới hạn sau

1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)

2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)

b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)

c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)

2

NV

Nguyễn Việt Lâm

Giáo viên

17 tháng 1 2021

\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)

\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)

DT

Dương thị bầu

15 tháng 3 2022

Lim 3.4n-2.13n/5n+6.13n

NN

Núi non tình yêu thuần khiết

15 tháng 2 2022

Tìm các giới hạn sau:

\(a,\dfrac{\left(2n+1\right)\left(3n-2\right)^2}{n^3+n-1}\)

\(b,\dfrac{2n-1}{3n^2+4n-1}\)

1

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 2 2022

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

\(=\lim\dfrac{\left(2+\dfrac{1}{n}\right)\left(3-\dfrac{2}{n}\right)^2}{1+\dfrac{1}{n^2}-\dfrac{1}{n^3}}=\dfrac{2.3^2}{1}=18\)

\(\lim\dfrac{2n-1}{3n^2+4n-1}=\lim\dfrac{n\left(2-\dfrac{1}{n}\right)}{n^2\left(3+\dfrac{4}{n}-\dfrac{1}{n^2}\right)}=\lim\dfrac{2-\dfrac{1}{n}}{n\left(3+\dfrac{4}{n}-\dfrac{1}{n^2}\right)}=\dfrac{2}{+\infty}=0\)