Cho B=3+3^2+3^3+...+3^2014+3^2015.Tìm x để 2B+3=3x
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B = 3 + 32 + 33 + ... + 32014 + 32015
3B = 3( 3 + 32 + 33 + ... + 32014 + 32015 )
3B = 32 + 33 + ... + 32015 + 32016
2B = 3B - B
= 32 + 33 + ... + 32015 + 32016 - ( 3 + 32 + 33 + ... + 32014 + 32015 )
= 32 + 33 + ... + 32015 + 32016 - 3 - 32 - 33 - ... - 32014 - 32015
= 32016 - 3
2B + 3 = 3x
<=> 32016 - 3 + 3 = 3x
<=> 32016 = 3x
<=> x = 2016
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{2014}+3^{2015}\)
\(3B=3^2+3^3+3^4+...+3^{2015}+3^{2016}\)
\(3B-B=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+...+3^{2015}\right)\)
\(2B=3^{2016}-3\). \(\Leftrightarrow3^{2016}-1+3=3^x\Leftrightarrow3^{2016}=3^x\Leftrightarrow x=2016\)
Vậy \(x=2016\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{2014}+3^{2015}\)
=>\(3B=3^2+3^3+3^4+...+3^{2015}+3^{2016}\)
=>\(3B-B=3^2+3^3+3^4+...+3^{2015}+3^{2016}-3-3^2-3^3-...-3^{2014}-3^{2015}\)
=>\(2B=3^{2016}-3\)
=>\(2B+3=3^{2016}\) là lũy thừa của 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$B=3+3^2+3^3+...+3^{2014}+3^{2015}$
$3B=3^2+3^3+3^4+....+3^{2015}+3^{2016}$
$\Rightarrow 2B=3B-B=3^{2016}-3$
$\Rightarrow 2B+3=3^{2016}$ là lũy thừa của $3$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(S=1+3+3^2+....+3^{2014}\)
\(\Rightarrow\left(3-1\right)A=\left(3-1\right)1+\left(3-1\right)3+\left(3-1\right)3^2+....+\left(3-1\right)3^{2014}\)
\(\Rightarrow2A=3-1+3-3^2+....+3^{2015}-3^{2014}\)
\(\Rightarrow2A=3^{2015}-1\)
\(\Rightarrow2B-2A=3^{2015}-\left(3^{2015}-1\right)\)
\(\Rightarrow2B-2A=1\)
\(\Rightarrow2\left(B-A\right)=1\)
\(\Rightarrow B-A=\frac{1}{2}\)
S = 1 + 3 + 32 + ... + 32014
= > ( 3 - 1 ) A = ( 3 - 1 ) 1 + ( 3 - 1 ) 3 + ( 3 - 1 ) 32 + ... + ( 3 - 1 ) 32014
= > 2A = 3 - 1 + 3 - 32 + ... + 32015 - 32014
= > 2A = 32015 - 1
= > 2B - 2A = 32015 - ( 32015 - 1 )
= > 2B - 2A = 1
= > 2 ( B - A ) = 1
= > B - A = \(\frac{1}{2}\)
Vậy B - A = \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có B = 3 + 32 + 33 + ... + 32014 + 32015
=> 3B = 32 + 33 + 34 + .... + 32015 + 32016
Lấy 3B trừ B theo vế ta có
3B - B = (32 + 33 + 34 + .... + 32015 + 32016) - (3 + 32 + 33 + ... + 32014 + 32015)
2B = 32016 - 3
Khi đó 2B + 3 = 3x
<=> 32016 - 3 + 3 = 3x
=> 32016 = 3x
=> x = 2016
Vậy x = 2016
Bg
Ta có: B = 3 + 32 + 33 +...+ 32014 + 32015
=> 3B = 3.(3 + 32 + 33 +...+ 32014 + 32015)
=> 3B = 3.3 + 3.32 + 3.33 +...+ 3.32014 + 3.32015
=> 3B = 32 + 33 + 34 +...+ 32015 + 32016
=> 3B - B = (32 + 33 + 34 +...+ 32015 + 32016) - (3 + 32 + 33 +...+ 32014 + 32015)
=> 2B = 32016 - 3
2B + 3 = 3x
=> 32016 - 3 + 3 = 3x
=> 32016 = 3x
=> x = 2016
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{2014}+3^{2015}\)
=>\(3B=3^2+3^3+3^4+...+3^{2015}+3^{2016}\)
=>\(3B-B=3^2+3^3+...+3^{2015}+3^{2016}-3-3^2-3^3-...-3^{2014}-3^{2015}\)
=>\(2B=3^{2016}-3\)
=>\(2B+3=3^{2016}\) là lũy thừa của 3
\(B+1=3^{2015}+3^{2014}+...+3^3+3^2+3+1\)
\(\Leftrightarrow2\left(B+1\right)=\left(3-1\right)\left(3^{2015}+3^{2014}+...+3^3+3^2+3+1\right)\)
\(\Leftrightarrow2B+2=3^{2016}-1\Leftrightarrow2B+3=3^{2016}\)
Vậy để \(2B+3=3^x\)thì x = 2016.