Tính :a, \(\frac{\frac{\left(-5\right)^n}{\left(7\right)}}{\frac{\left(-5\right)^{n-1}}{7}}\left(n>=1\right)\) b,\(\frac{\frac{\left(-1\right)^{2n}}{2}}{\frac{\left(-1\right)^n}{2}}\left(n\in...

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NV

Nguyễn Văn Duy

27 tháng 3 2017

Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

2

NN

Ngu Ngu Ngu

27 tháng 3 2017

Ta có:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)

\(=1-\frac{2n+1}{\left(n+1\right)^2}\)

Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)

VX

Vũ Xuân Phương

28 tháng 3 2017

SAI RỒI ĐÁP ÁN LÀ N^2/(N+1)^2

NP

Nguyễn Phương Nhi

8 tháng 10 2016

Tính :

a) \(\frac{\left(\frac{-5}{7}\right)^n}{\left(\frac{-5}{7}\right)^{n-1}}\)( n\(\ge\)1 )

b) \(\frac{\frac{-1}{2}^{2n}}{\left(\frac{-1}{2}\right)^n}\) ( n \(\in\)N )

1

NL

Nguyễn Lê Phước Thịnh

4 tháng 2 2022

a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)

b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)

1. Chứng minh : B = \(\left(1-\frac{2}{6}\right).\left(1-\frac{2}{12}\right).\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)2. cho M = \(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+\frac{1}{5.\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)N = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}\)Rút...

TT

Thanh Tùng DZ

4 tháng 10 2017

0

LC

LeO Channel

16 tháng 8 2017

Chứng minh: \(\frac{3}{\left(1x2\right)}+\frac{5}{\left(2x3\right)}+...+\frac{2n+1}{\left(n\left(n+1\right)\right)^2}=\frac{n\left(n+2\right)}{\left(n+1\right)^2}\)

0

T

Tuấn

25 tháng 2 2017

help me! (ngu toàn...

1

LD

l҉o҉n҉g҉ d҉z҉

26 tháng 2 2017

\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)

\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)

\(=50.\frac{9}{50}=9\)

Tính :1 + 3 + 5 + 7 + ... + (2n - 1) = 225 Giải :Theo công thức tính dãy số , ta có :\(\frac{\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}.\left[\left(2n-1\right)+1\right]}{2}=225\)\(\frac{\left\{\left[2n-2\right]:2+1\right\}.2n}{2}=225\)\(\left\{\left[2n-2\right]:2+1\right\}.n=450\)(Lượt giản thừa số 2)\(\left\{\frac{2n-2}{2}+1\right\}.n=225\)\(\left\{\frac{2n-2}{2}+\frac{2}{2}\right\}.n=225\)\(\frac{2n-2+2}{2}.n=225\)\(\frac{2n}{2}.n=225\)\(n^2=225\)\(\Rightarrow...

QC

Quay Cuồng

16 tháng 8 2017

CMR \(\forall n\in\)N* ta có

\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)

#Toán lớp 6

0

TD

Tín Đinh

2 tháng 7 2017

Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)

#Toán lớp 9

0

KR

Kayasari Ryuunosuke

1 tháng 12 2016

Giúp mik vớiTính nhanh:a. A=\(\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\left(n\in N\right)\)b. B=\(\left(10000-1^2\right)\left(10000-2^2\right)\left(10000-3^2\right)..\left(10000-1000^2\right)\)c. C=\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)d....

0

BP

Bùi Phúc Hoàng Linh

21 tháng 8 2020

1

NN

Nobi Nobita

21 tháng 8 2020

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

b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)

c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)

\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)