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\(A=3+3^2+3^3+...+3^{100}\)
\(3A=3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-3\)
\(A=\left(3^{101}-3\right):2\)
Ta có : \(2A+3=3^{101}\)
\(→n=101\)
~ Ủng hộ nhé ~
a) 2^n=128/4=32=2^5\(\Rightarrow\)n=5
b)3^n+1 :9=81\(\Rightarrow\)3^n.3 :9=81\(\Rightarrow\)3^n:3=81\(\Rightarrow\)3^n =243=3^5\(\Rightarrow\)n=5
c) 15^n:15=(3^2)^2:3^4=3^4:3^4=1\(\Rightarrow\)15^n=15=15^1\(\Rightarrow\)n=1
a, <=> 2^n = 128/4 = 32
<=> 2^n = 2^5
<=> n =5
b,<=> 3^(n+1) = 81.9= 729
<=> 3^(n+1) = 3^6
<=> n+1 = 6 <=> n =5
c, <=> 15^(n-1) = 1
<=> 15^(n-1) = 15^ 0
<=> n-1 = 0 <=> n =1
a)3.2n=48
<=>2n=48:3
<=>2n=16
<=>2n=24
=>n=4
b)64.16n=1024
<=>16n=1024:64
<=>16n=16
<=>16n=161
=>n=1
c)5n.5n=580
<=>52n=52x40
=>n=40
d)5n+1-5n=500
<=>5n.5-5n=500
<=>5n.(5-1)=500
<=>5n.4=500
<=>5n=125
=>n=3
a) \(3.2^n=48\\ 2^n=16\\ 2^n=2^4\\ n=4\)
a) 32 < 2n > 128
<=> 25 < 2n > 27
<=> n = 8 ; 9 ; 10...
b) 2 . 16 < 2n > 4
<=> 21 . 24 < 2n > 4
<=> 25 < 2n > 4
<=> n = 5 ; 6 ; 7 ;...
c) ( 22 : 4 ) . 2n = 4
<=> 1 . 2n = 4
<=> 2n = 4
<=> 2n = 22
<=> n = 2
\(n^2+\left(n+1\right)^2+\left(n+2\right)^2=\left(n+5\right)^2\)
\(\Leftrightarrow n^2+n^2+2n+1+n^2+4n+4=n^2+10n+25\)
\(\Leftrightarrow2n^2-4n-20=0\)
\(\Leftrightarrow2\left(n^2-2n-10\right)=0\)
\(\Leftrightarrow n^2-2n-10=0\)
\(\Leftrightarrow n^2-2n+1-11=0\)
\(\Leftrightarrow\left(n-1\right)^2-11=0\)
\(\Leftrightarrow\left(n-1-\sqrt{11}\right)\left(n-1+\sqrt{11}=0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}n-1-\sqrt{11}=0\\n-1+\sqrt{11}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=\sqrt{11}+1\\n=-\sqrt{11}+1\end{cases}}\)
Vậy: \(S=\left\{\sqrt{11}+1;-\sqrt{11}+1\right\}\)
\(32^n+16^n=1024\)
\(\Leftrightarrow\left(2^5\right)^n+\left(2^4\right)^n=2^{10}\)
\(\Leftrightarrow2^{5n}.2^{4n}=2^{10}\)
\(\Leftrightarrow2^{4n+5n}=2^{10}\)
\(\Leftrightarrow9n=10\Leftrightarrow n=\frac{10}{9}\)