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Ta có : 3A = 32 + 33 + 34 + 35 + .... + 32010
=> 3A - A = 32010 - 3
=> 2A = 32010 - 3
Ta có : 2A + 3 = 3n
=> 32010 - 3 + 3 = 3n
=> 32010 = 3n
=> n = 2010
vậy n = 2010
=>3A=32+33+…+32010
=>3A-A=32+33+…+32010-3-32-…-32009
=>2A=32010-3
=>2A+3=32010=3N
=>N=2010
A = 3+32+33+......+32009
3A = 32+33+34+......+32010
2A = 3A - A = 32010-3
=> 2A + 3 = 32010
Mà 2A + 3 = 3n
=> n = 2010
3A=3^2+3^3+3^4+...+3^2010
2A=3^2010-3
2A+3=3^2010-3+3=3^n
3^2010=3^n
n=2010
A=3+3^2+3^3+...+3^2009
=>3A=3^2+3^3+3^4+...+3^2010
=>3A-A=3^2010-3
=>2A=3^2010-3
=>2A+3=3^2010
=>n=2010
\(3A=3^2+3^3+3^4+...+3^{2010}\)
\(3A-A=\left(3^2+3^3+3^4+..+3^{2010}\right)-\left(3+3^2+3^3+....+3^{2009}\right)\)
\(2A=3^{2010}-3\)(1)
(1) => \(3^{2010}-3+3=3^{2010}\)
=> n = 2010
A = 3 + 32 + 33 + ... + 32009
3A = 32 + 33 + 34 + ... + 32010
3A - A = (32 + 33 + 34 + ... + 32010) - (3 + 32 + 33 + ... + 32009)
2A = 32010 - 3
3n = 2A + 3
3n = 22010 - 3 + 3
3n = 32010
n = 2010
A=1+3^1+3^2+...+3^2008
3A=3(1+3^1+3^2+...+3^2008)
3A=3*1+3*3^1+3*3^2+...+3*3^2008
3A=3+3^2+3^3+...+3^2009
3A-A=(3+3^2+3^3+...+3^2009)-(1+3^1+3^2+...+3^2008)
A=(3^2009-1):2
=>2A=(3^2009-1):2
<=>A=3^2009-1
vi 2 so lien tiep hon kem nhau 1 don vi
=>3^2009-1 va 3^2009 la 2 so lien tiep
=>2A va B la 2 so tu nhien lien tiep
\(A=3+3^2+3^3+...+3^{120}\)
\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{100}\right)\)
\(3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^3+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}-3+3=3^{101}=3^n\)
\(\Rightarrow n=101\)
vậy ...
a, \(3^{-2}.3^4.3^3.3^n=3^7\)
\(\Rightarrow3^{-2+4+3+n}=3^7\)
\(\Rightarrow3^{5+n}=3^7\)
Vì \(3\ne-1;3\ne0;3\ne1\) nên \(5+n=7\Rightarrow x=2\)
Vậy....
b, \(2^{-1}.2^n+4.2^n=9.2^5\)
\(\Rightarrow2^n.\left(2^{-1}+2^2\right)=9.2^5\)
\(\Rightarrow2^n.4,5=9.32\)
\(\Rightarrow2^n=288:4,5\)
\(\Rightarrow2^n=64=2^6\)
Vì \(2\ne-1;2\ne0;2\ne1\) nên \(n=6\)
Vậy.....
c, \(2.16\ge2^n>4\)
\(\Rightarrow32\ge2^n>2^2\)
\(\Rightarrow2^5\ge2^n>2^2\)
Vì \(2\ne-1;2\ne0;2\ne1\) nên \(5\ge n>2\)
\(\Rightarrow n\in\left\{3;4;5\right\}\)
Vậy.....
Chúc bạn học tốt!!!
Câu hỏi của nguyen lan anh - Toán lớp 7 | Học trực tuyến
Đề : Cho A= 1+3+32+33+34+...+32000. Biết 2A=3n-1
Tìm n
3A=\(3^2+3^3+...+3^{2010}\)
\(3A-A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+....+\left(3^{2009}-3^{2009}\right)+3^{2010}-3\)
\(2A=3^{2010}-3\)
\(A=\frac{3^{2010}-3}{2}\)
\(2A+3=\frac{3^{2010}-3}{2}\times2+3=3^{2010}-3+3=3^{2010}\)
Mà 32010 = 3n
Nên n = 2010