a) \(\dfrac{1}{27}\cdot81^n=3^n\)
\(\Rightarrow\dfrac{1}{3^3}\cdot3^{4n}=3^n\)
\(\Rightarrow\dfrac{3^{4n}}{3^3}=3^n\)
\(\Rightarrow3^{4n-3}=3^n\)
\(\Rightarrow4n-3=n\)
\(\Rightarrow3n=3\)
\(\Rightarrow n=1\)
b) Ta có:
\(C=3+3^2+3^3+...+3^{100}\)
\(3C=3^2+3^3+...+3^{101}\)
\(3C-C=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
\(2C=3^{101}-3\)
\(C=\dfrac{3^{101}-3}{2}\)
Nên ta có:
\(2C+3=3^n\)
\(\Rightarrow2\cdot\dfrac{3^{101}-3}{2}+3=3^n\)
\(\Rightarrow3^{101}-3+3=3^n\)
\(\Rightarrow3^{101}=3^n\)
\(\Rightarrow n=101\)
c) \(2\cdot16\ge2^n>4\)
\(\Rightarrow2\cdot2^4\ge2^n>2^2\)
\(\Rightarrow2^5\ge2^n>2^2\)
\(\Rightarrow5\ge n>2\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`1/27 * 81^n = 3^n`
`=> 1/27 = 3^n \div 81^n`
`=> 1/27 = 3^n \div 3^(4n)`
`=> 1/27 = 3^(n - 4n)`
`=> 3^(-3) = 3^(-3n)`
`=> -3n = -3`
`=> 3n = 3`
`=> n = 1`
Vậy, `n = 1`
`b)`
`C = 3+ 3^2 + 3^3 + ... + 3^100`
`=> 3C = 3^2 + 3^3 + 3^4 + ... + 3^101`
`=> 3C - C = (3^2 + 3^3 + 3^4 + ... + 3^101) - (3 + 3^2 + 3^3 + ... + 3^100)`
`=> 2C = 3^2 + 3^3 + 3^4 + ... + 3^101 - 3 - 3^2 - 3^3 - ... - 3^100`
`=> 2C = 3^101 - 3`
`2C + 3 = 3^n`
`=> 3^101 - 3 + 3 = 3^n`
`=> 3^101 = 3^n`
`=> n = 101`
Vậy, `n= 101`
`c)`
`2*16 \ge 2^n > 4`
`=> 2*2^4 \ge 2^n > 2^2`
`=> 2^5 \ge 2^n > 2^2`
`=> n \in {3; 4; 5}.`
a) \(\dfrac{1}{27}.81^n=3^n\)
\(\Rightarrow\dfrac{3^{4n}}{3^3}=3^n\)
\(\Rightarrow3^{4n-3}=3^n\)
\(\Rightarrow4n-3=n\)
\(\Rightarrow4n-n=3\Leftrightarrow n=1\)
b) \(C=3+3^2+3^3+...+3^{100}\)
\(3.C=3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(3C-C=3^2+3^3+3^4+...+3^{100}+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\)
\(2C=3^{101}-3\)
\(\Rightarrow2C+3=3^{101}\)
Mặt khác: \(2C+3=3^n\) \(\Rightarrow n=101\)
c) \(2.16\ge2^n>4\)
\(\Rightarrow2.2^4\ge2^n>2^2\)
\(\Rightarrow2^5\ge2^n>2^2\Leftrightarrow n\in\left\{3;4;5\right\}\) (do n là số tự nhiên)
