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\(3A=3^2+3^3+3^4+...+3^{2021}\)
\(2A=3A-A=3^{2021}-3\)
\(\Rightarrow2A+3=3^{2021}-3+3=3^{2021}=3^n\Rightarrow n=2021\)
Ta có: 3A=32+33+...+3101
3A-A=2A=(32+33+...+3101)-(3+32+...+3100)
2A=3101-3
A=\(\frac{3^{101}-3}{2}\)
=>2A+3=2.\(\frac{3^{101}-3}{2}\)+3
=(3101-3)+3
=3101
Mà 2A+3=3n
=>3101=3n
=>n=101
A=3+32+33+...+3100
2A=(3+32+33+...+3100)x2
2A=32+33+34...+3101
2A-A=3101-3
mà 3n=2A+3=3101-3+3=3101
suy ra n=101
\(A=1+3+3^2+...+3^{2016}+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2017}+3^{2018}\)
\(3A-A=3^{2018}-1\)
\(2A+1=3^{2018}\)
Vậy n = 2018
3A=3+3^2+3^3+...+3^2018
-A=1+3+3^2+...+3^2017
2A=3^2018-1
khi đó ta có 2A+1=3^2018-1+1=3^2018=3^n
=>n=2018
A=\(3+3^2+3^3+...+3^{100}\)
3A=\(3^2+3^3+3^4+...+3^{101}\)
3A - A=\(3^2+3^3+3^4+...+3^{101}-3-3^2-3^3-...-3^{100}\)
2A = \(3^{101}-3\)
=>\(2A+3=3^n\)
=>\(3^{101}-3+3=3^n\)
=>3\(^{101}=3^n\)
=>n=101
A=3+3^2+3^3+...+3^100
=>3A=3^2+3^3+3^4+...+3^101
=>3A-A=2A=3^101-3
mà 2A+3=3^n
=>3^101-3+3=3^n
=>3^n=3^101
=>n=101
A=3+32+33+...+3100
3A=32+33+...+3101
3A-A=(32+33+...+3101)-(3+32+33+...+3100)
2A=3101-3
2A+3=3101
\(A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3.\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3A-A=2A=\left[3^2+3^3+3^4+...+3^{101}\right]-\left[3+3^2+3^3+...+3^{100}\right]\)\(\Rightarrow2A=3^{101}-3\)
Theo đề bài ta có 2A + 3 = 3n ( \(n\in N\) )
\(\Rightarrow2A+3=3^{101}-3+3=3^n\)
\(\Rightarrow2A+3=3^{101}=3^n\)
\(\Rightarrow3^{101}=3^n\)
\(\Rightarrow101=n\) ( thỏa mãn điều kiện \(n\in N\)
Vậy n = 101
A = 3 + 32 + 33 + ...+3100
3A = 32 + 33 + 34 + ...+ 3101
3A - A = ( 32 + 33 + 34 + ...+ 3101 ) - ( 3 + 32 + 33 + ...+3100 )
2A = 3101 - 3
Thay vào 2A + 3 = 3n ta có
3101 - 3 + 3 = 3n
3101 = 3n
=> n = 101
A = 3 + 32 + 33 +....+ 3100
\(\Rightarrow\) 3A= 3.(3 + 32 + 33 +....+ 3100)
\(\Rightarrow\) 3A= 32 + 33 + 34 +.....+ 3101
\(\Rightarrow\)3A - A= (32 + 33 + 34 +.....+ 3101) - (3 + 32 + 33 +....+ 3100)
\(\Rightarrow\)2A= 3101 - 3
mà 2A + 3 = 3n
\(\Rightarrow\)3101 - 3 + 3 = 3n
\(\Rightarrow\)3101 = 3n
\(\Rightarrow\)n=101