HK
1
K
Khách
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NT
0
A
8 tháng 3 2019
\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-1+\frac{1}{100}\)
\(A=\frac{2}{100}-1\)
\(A=\frac{1}{50}-1\)
\(A=\frac{-49}{50}\)
A
8 tháng 3 2019
b,\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\) (1)
Đặt \(B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\)
\(\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)\)
\(=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}\)
\(2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)\)
\(=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)\)
\(B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2\)
\(=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3\)
Đặt \(C=2^3+2^4+2^5+2^n\)
\(\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)\)
\(C=2^4+2^5+2^6+...+2^{n+1}\)
\(2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)\)
\(C=2^{n+1}-2^3\)
Khi đó : \(B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3\)
\(=-2^{n+1}+2^3+n.2^{n+1}-2^3\)
=\(=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}\)
Vậy từ (1) ta có:\(\left(n-1\right),2^{n+1}=2^{n+34}\)
\(2^{n+34}-\left(n-1\right).2^{n+1}=0\)
\(2^{n+1}.[2^{33}-\left(n-1\right)]=0\)
Do đó \(2^{33}-n+1=0\)( Vì \(2^{n+1}\ne0\)với mọi \(n\))
\(n=2^{33}+1\)
Vậy \(n=2^{33}+1\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1
Đặt \(A=2.2^2+3.2^3+...+n.2^n\)
\(\Rightarrow2A=2.2^3+3.2^4+...+n.2^{n+1}\)
\(\Rightarrow A-2A=2.2^2+\left(3.2^3-2.2^3\right)+...+\left[n.2^n-\left(n-1\right).2^n\right]-n.2^{n-1}\)
\(\Rightarrow-A=2.2^2+2^3+2^4+...+2^n-n.2^{n+1}\)
\(\Rightarrow-A=2+2^1+2^2+2^3+...+2^n-n.2^{n+1}\)
\(\Rightarrow-2A=4+2^2+2^3+...+2^{n+1}-n.2^{n+2}\)
\(\Rightarrow-A-\left(-2A\right)=2+2^1-4-n.2^{n+1}-2^{n+1}+n.2^{n+2}\)
\(\Rightarrow A=n.2^{n+2}-\left(n+1\right)2^{n+1}\)
\(\Rightarrow A=2n.2^{n+1}-\left(n+1\right)2^{n+1}\)
\(\Rightarrow A=\left(n-1\right).2^{n+1}\)
NM
1
MH
17 tháng 10 2023
\(Đặt\) \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
\(2A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}\)
\(2A-A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}-\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)
\(=-2.2^2-2^3-2^4-...-2^n+n.2^{n+1}\)
\(=-2^2-\left(2^2+2^3+...+2^n\right)+n.2^{n+1}\)
\(=-2^2-\left(2^{n+1}-2^2\right)+n.2^{n+1}\)
\(=\left(n-1\right).2^{n+1}\)
=> \(\left(n-1\right).2^{n+1}=2^{n+16}=2^{n+1}.2^{15}\)
\(\Leftrightarrow n-1=2^{15}\)
\(\Leftrightarrow n=2^{15}+1\)
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Lời giải:
$2^n+34=2.2^2+3.2^3+....+n.2^n$
$2^{n+1}+68=2.2^3+3.2^4+....+n.2^{n+1}$
Trừ theo vế:
$2^n+34=n.2^{n+1}-(8+2^3+2^4+...+2^n)$
$n.2^{n+1}-2^n-42=2^3+2^4+...+2^n$
$n.2^{n+2}-2^{n+1}-84=2^4+....+2^{n+1}$
Trừ theo vế:
$n.2^{n+1}-2^n-42=2^{n+1}-8$
$2^n(2n-3)=34=17.2$
$\Rightarrow 2^n=2$ và $2n-3=17$ (vô lý)
Vậy không tìm được $n$.
\(2A=2.2^3+3.2^4+4.2^5+...+100.2^{101}\)
=> \(2A-A=100.2^{101}-\left(2^{100}+2^{99}+...+2^4+2^3\right)-2.2^2\)
Đặt \(B=2^3+2^4+...+2^{100}\Rightarrow2B=2^4+2^5+...+2^{101}\)
=> \(2B-B=2^{101}-2^3\Rightarrow B=2^{101}-2^3\)
=> \(2A-A=100.2^{101}-\left(2^{101}-2^3\right)-2.2^2\)
=> \(A=\left(100.2^{101}-2^{101}\right)+2^3-2^3\)=\(99.2^{101}\)