A=4+4²+4³+...+4¹⁹+4²⁰
B=1+3+3²+3³+...3⁹⁹
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a)Ta có :
\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)
\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)
\(3A=1-\dfrac{1}{4^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)
~ Chúc bn học tốt ~
a: \(\dfrac{3}{4}A=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+...+\left(\dfrac{3}{4}\right)^{2021}\)
=>\(\dfrac{7}{4}\cdot A=\left(\dfrac{3}{4}\right)^{2021}+1\)
=>\(A\cdot\dfrac{7}{4}=\dfrac{3^{2021}+4^{2021}}{4^{2021}}\)
=>\(A=\dfrac{3^{2021}+4^{2021}}{4^{2020}\cdot7}\)
b: Vì 3^2021+4^2021 ko chia hết cho 4^2020*7 nên A ko là số nguyên
1 - \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) = \(\dfrac{4}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) = \(\dfrac{2}{4}\) = \(\dfrac{1}{2}\)
Vậy a, S còn b, Đ
Ta có công thức tổng quát như sau:
\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)
Áp dụng ta có:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)
1)
a) \(-\frac{9}{34}:\frac{17}{4}\)
\(=-\frac{18}{289}.\)
b) \(1\frac{1}{2}.\frac{1}{24}\)
\(=\frac{3}{2}.\frac{1}{24}\)
\(=\frac{1}{16}.\)
c) \(-\frac{5}{2}:\frac{3}{4}\)
\(=-\frac{10}{3}.\)
d) \(4\frac{1}{5}:\left(-2\frac{4}{5}\right)\)
\(=\frac{21}{5}:\left(-\frac{14}{5}\right)\)
\(=-\frac{3}{2}.\)
Mấy câu sau bạn đăng ríu rít quá khó nhìn lắm.
Chúc bạn học tốt!
a, A = 4 + 4^2 + 4^3 + ... + 4^n
=> 4A = 4.(4 + 4^2 + 4^3 + ... + 4^n)
=> 4A = 4^2 + 4^3 + 4^4 + ... + 4^n+1
=> 3A = 4A - A = (4^2 + 4^3 + 4^4 + ... + 4^n+1) - ( 4 + 4^2 + 4^3 + ... + 4^n)
=> 3A = 4^n+1 - 4
=> A = \(\frac{4^{n+1}-4}{3}\)
Vậy A = ..................
b, B = 1 + 3 + 3^2 + ... + 3^100
=> 3B = 3.(1 + 3 + 3^2 + ... + 3^100)
=> 3B = 3 + 3^2 + 3^3 + ... + 3^101
=> 2B = 3B - B =(3 + 3^2 + 3^3 + ... + 3^101) -(1 + 3 + 3^2 + ... + 3^100)
=> 2B = 3^101 - 1
=> B = \(\frac{3^{101}-1}{2}\)
Vậy B = ......................
A = 4 + 4^2 + 4^3 + ... + 4^n
4A = 4^2 + 4^3 + 4^4 + ... + 4^n+1
4A - A = ( 4^2 + 4^3 + 4^4 + ... + 4^n+1 ) - ( 4 + 4^2 + 4^3 +...+4^n)
3A = 4^n+1 - 4
A = 4^n+1 - 4/3
a) A= 1/3 – 3/4 – (-3/5) + 1/15 – 2/9 – 1/36 b) 2/5 + (-4/3) - 1/2 c) 1/3 – [( -5/4) – (1/4 + 3/8)]
Bài 1
a) 3 2/5 - 1/2
= 17/5 - 1/2
= 34/10 - 5/10
= 29/10
b) 4/5 + 1/5 × 3/4
= 4/5 + 3/20
= 16/20 + 3/20
= 19/20
c) 3 1/2 × 1 1/7
= 7/2 × 8/7
= 4
d) 4 1/6 : 2 1/3
= 25/6 : 7/3
= 25/14
Bài 2
a) 3 × 1/2 + 1/4 × 1/3
= 3/2 + 1/12
= 18/12 + 1/12
= 19/12
b) 1 4/5 - 2/3 : 2 1/3
= 9/5 - 2/3 : 7/3
= 9/5 - 2/7
= 63/35 - 10/35
= 53/35
\(A=4+4^2+4^3+...+4^{19}+4^{20}\)
\(4A=4\cdot\left(4+4^2+4^3+...+4^{20}\right)\)
\(4A=4^2+4^3+...+4^{21}\)
\(4A-A=4^2+4^3+4^4+...+4^{21}-4-4^2-4^3-...-4^{20}\)
\(3A=4^{21}-4\)
\(A=\dfrac{4^{21}-4}{3}\)
____________
\(B=1+3+3^2+...+3^{99}\)
\(3B=3\cdot\left(1+3+3^2+....+3^{99}\right)\)
\(3B=3+3^2+3^3+....+3^{100}\)
\(3B-B=\left(3+3^2+3^3+....+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\)
\(2B=3^{100}-1\)
\(B=\dfrac{3^{100}-1}{2}\)
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