cho f(n)=(n2 n 1 )2 1 với n thuộc N* . Đặt \(p_n=\frac{f_{\left(1\right)}\cdot f_{\left(3\right)}\cdot f_{\left(5\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n-1\right)}}{f

TT

Trần Thị Hà Phương

8 tháng 10 2017

cho f(n)=(n2 + n +1 )2 +1 với n thuộc N* . Đặt \(p_n=\frac{f_{\left(1\right)}\cdot f_{\left(3\right)}\cdot f_{\left(5\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n-1\right)}}{f_{\left(2\right)}\cdot f_{\left(4\right)}\cdot f_{\left(6\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n\right)}}\)chứng minh rằng : P1 + P2 +P3 +................+...

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#Toán lớp 9

0

PT

Pham Tien Dat

27 tháng 12 2020

\(f\left(n\right)=\left(n^2+n+1\right)^2+1\). Xét dãy \(\left(u_n\right)\) sao cho : \(\left(u_n\right)=\dfrac{f\left(1\right)\cdot f\left(3\right)\cdot f\left(5\right)...\cdot f\left(2n-1\right)}{f\left(2\right)\cdot f\left(4\right)\cdot...\cdot f\left(2n\right)}\). Tính \(\lim\limits_{n\sqrt{u_n}}\)

#Toán lớp 11

1

YS

yr shio

27 tháng 12 2020

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TM

Trần Mai Anh

12 tháng 2 2017

Chứng minh rằng

\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)\cdot...\cdot2n}=\frac{1}{2^n}\)

#Toán lớp 6

0

O

online

12 tháng 4 2018

Chứng minh rằng:

a)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)

b)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\cdot\cdot\cdot2n}=\frac{1}{2^n}\)Với \(n\inℕ^∗\)

#Toán lớp 6

0

FC

Five centimeters per second

28 tháng 2 2017

Chứng minh rằng :

\(\frac{1\cdot3\cdot5\cdot...\cdot\left(2n-1\right)}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)\cdot...\cdot2n}=\frac{1}{2^n}\)

#Toán lớp 6

0

NT

Nguyễn Thị Thanh Dung

17 tháng 9 2020

Tính các tích sau: với n là số tự nhiên, n<3

a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{n}\right)\)

b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)

#Toán lớp 7

0

P

phamngocson

25 tháng 6 2016

TÍNH...

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#Toán lớp 6

1

DT

Đinh Thùy Linh

26 tháng 6 2016

\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)

\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)

Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:

\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)

\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)

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WT

Where there is love there is life

25 tháng 7 2018

Tính:

\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)

#Toán lớp 10

1

NL

Nguyễn Lê Phước Thịnh

29 tháng 8 2022

\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)

\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)

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NP

Nguyễn Phong

19 tháng 6 2019

Thu gọn biểu thức sau :

a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)

b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)

#Toán lớp 8

2

PT

Phạm Thị Thùy Linh

19 tháng 6 2019

\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=2^{64}-1-2^{64}=-1\)

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DL

Duc Loi

19 tháng 6 2019

a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)

Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)

\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)

b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.

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YY

yo yo

12 tháng 6 2018

Chứng minh rằng với số nguyên dương \(n\ge6\) thì số

\(a_n=1+\dfrac{2\cdot6\cdot10\cdot\cdot\cdot\left(4n-2\right)}{\left(n+5\right)\left(n+6\right)\cdot\cdot\cdot\left(2n\right)}\) là số chính phương

#Toán lớp 10

0

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