Cho B=1+3+32+...+3100
a) Tính Tổng B
b)Tìm x, biết rằng 2B+1=3
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a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
\(A=3+3^2+3^3+...+3^{100}\)
=>\(3\cdot A=3^2+3^3+...+3^{101}\)
=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2A=3^{101}-3\)
=>\(2A+3=3^{101}\)
mà \(2A+3=3^n\)
nên n=101
b: \(A=3+3^2+...+3^{100}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{98}\right)⋮12\)
1.
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{4}$
$=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=\frac{-20}{20}=-1$
$\Rightarrow a=2(-1)=-2; b=3(-1)=-3; c=4(-1)=-4$
2.
$S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}$
$=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
a) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right)\cdot2=-2\\b=\dfrac{\left(-1\right).6}{2}=-3\\c=\dfrac{\left(-1\right).12}{3}=-4\end{matrix}\right.\)
b) Ta có : \(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\).
Vậy : \(S=\dfrac{99}{100}.\)
a)\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=-\dfrac{20}{20}=-1\)
\(\left\{{}\begin{matrix}\dfrac{a}{2}=-1\Leftrightarrow a=-2\\\dfrac{b}{3}=-1\Leftrightarrow b=-3\\\dfrac{c}{4}=-1\Leftrightarrow c=-4\end{matrix}\right.\)
b)\(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
1/ Vì x,y,z tỉ lệ với 3,5,7 nên \(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=\frac{x-z}{3-7}=\frac{12}{-4}=-3\)
=> x/3 = -3 => x = -9
y/5 = -3 => y = -15
z/7 = -3 => z = -21
Vậy x=-9,y=-15,z=-21
2/
Ta có: 3x = 4y => x/4 = y/3
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{21}{7}=3\)
=> x/4 = 3 => x = 12
y/3 = 3 => y = 9
Vậy x=12,y=9
3/
Ta có: 5a = 2b => a/2 = b/5 => 3a/6 = 2b/10
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{3a}{6}=\frac{2b}{10}=\frac{3a+2b}{6+10}=\frac{32}{16}=2\)
=> a/2 = 2 => a = 4
b/5 = 2 => b = 10
Vậy a=4,b=10
1.
a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)
\(2A=2+2^2+2^3+....+2^{2008}\)
b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)
\(=2^{2008}-1\) (bạn xem lại đề)
2.
\(A=1+3+3^1+3^2+...+3^7\)
a. \(2A=2+2.3+2.3^2+...+2.3^7\)
b.\(3A=3+3^2+3^3+...+3^8\)
\(2A=3^8-1\)
\(=>A=\dfrac{2^8-1}{2}\)
3
.\(B=1+3+3^2+..+3^{2006}\)
a. \(3B=3+3^2+3^3+...+3^{2007}\)
b. \(3B-B=2^{2007}-1\)
\(B=\dfrac{2^{2007}-1}{2}\)
4.
Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)
a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)
b.\(4C-C=4^7-1\)
\(C=\dfrac{4^7-1}{3}\)
5.
\(S=1+2+2^2+2^3+...+2^{2017}\)
\(2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(S=2^{2018}-1\)
4:
a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6
=>4*C=4+4^2+...+4^7
b: 4*C=4+4^2+...+4^7
C=1+4+...+4^6
=>3C=4^7-1
=>\(C=\dfrac{4^7-1}{3}\)
5:
2S=2+2^2+2^3+...+2^2018
=>2S-S=2^2018-1
=>S=2^2018-1
B=1+3+3^2+....+3^100
3B=3+3^2+3^3+...+3^101
3B-B=2B=(3+3^2+3^3+...+3^101)-(1+3+3^2+....+3^100)
2B= 3^101- 1
B=(3^101-1):2