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Đặt A = 3¹ + 3² + 3³ + 3⁴ + ... + 3⁹⁹ + 3¹⁰⁰
= (3¹ + 3²) + (3³ + 3⁴) + ... + (3⁹⁹ + 3¹⁰⁰)
= 3.(1 + 3) + 3³.(1 + 3) + ... + 3⁹⁹.(1 + 3)
= 3.4 + 3³.4 + ... + 3⁹⁹.4
= 4.(3 + 3³ + ... + 3⁹⁹) ⋮ 4
Vậy A ⋮ 4
Lời giải:
$B=3+(32+33+...+3100)$
$=3+\frac{(3100+32).3069}{2}=3+4806054=4806057$ không chia hết cho $160$
Bạn xem lại đề.
Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
\(A=3+3^2+3^3+3^4+.......+3^{100}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(\Rightarrow A=3.\left(1+3+3^2+3^3\right)+........+3^{97}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=3.40+.........+3^{97}.40\)
\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)
\(\Rightarrow A⋮40\)( 1 )
Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)
\(\Rightarrow A⋮120\)
Vậy \(A⋮120\)( ĐPCM )
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
\(B=3^1+3^2+3^3+...+3^{300}\\=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{299}+3^{300})\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+...+3^{299}\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{299}\cdot4\\=4\cdot(3+3^3+3^5+...+3^{299})\)
Vì \(4\cdot(3+3^3+3^5+...+3^{299})\vdots2\)
nên \(B\vdots2\)
B=(3+32)+(33+34)+...+(3299+3300)
B=3(1+3)+33(1+3)+...+3299(1+3)
B=3.4+33.4+...+3299.4
B=4(3+33+...+3299) chia hết cho 2 vì 4 chia hết cho 2
vậy B chia hết cho 2
A=2+22+23+...+299+2100A=2+22+23+...+299+2100
⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101
⇒A=2101−2⇒A=2101−2
B=3+32+33+...+399+3100B=3+32+33+...+399+3100
⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101
⇒2B=3101−3⇒2B=3101−3
⇒B=3101−32
\(B=3^1+3^2+3^3+...+3^{100}\)
\(B=\left(3^1+3^2\right)+...+\left(3^{99}+3^{100}\right)\)
\(B=3.\left(1+3\right)+...+3^{99}.\left(1+3\right)\)
\(B=3.4+...+3^{99}.4\)
\(B=4.\left(3^2+...+3^{99}\right)\)
Vì \(B⋮4\) nên \(B⋮2\left(đpcm\right)\)
Ta có B = 31 + 32 + 33 + ... + 3100
B = ( 31 + 32 ) + ( 33 + 34 ) + ... + ( 399 + 3100 )
B = 3( 1 + 3 ) + 33( 1 + 3 ) + ... + 399( 1 + 3 )
B = 3 . 4 + 33 . 4 + ... + 399 . 4
B = 4( 3 + 33 + ... + 399 ) ⋮ 2 vì 4 ⋮ 2
Vậy B ⋮ 2