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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:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)
=> 4A = 4 + 42 + 43 + ... + 42020
4A - A = 4 + 42 + ... + 42020 ) - ( 1 + 4 + ... + 42019 )
3A = 42020 - 1
A = \(\frac{4^{2020}-1}{3}\)
Ta có A - B = 0
Vậy A - B = 0
Ta có : A = 40 + 41 + 42 + .... + 42019
= 1+ 4 + 42 + .... + 42019
=> 4A = 4 + 42 + 43 + ... + 42020
Lấy 4A trừ A theo vế ta có :
\(4A-A=\left(4+4^2+4^3+...+4^{2020}\right)-\left(1+4+4^2+...+4^{2019}\right)\)
\(3A=4^{2020}-1\)
\(A=\frac{4^{2020}-1}{3}\)
\(\Rightarrow A-B=\frac{4^{2020}-1}{3}-\frac{4^{2020}}{3}=\frac{4^{2020}-1-4^{2020}}{3}=-\frac{1}{3}\)
\(A=1+4+4^2+4^3+...+4^{50}\)
\(4A=4\left(1+4+4^2+4^3+...+4^{50}\right)\)
\(4A=4+4^2+4^4+...+4^{51}\)
\(3A=4^{51}-1\)
\(A=\frac{\left(4^{51}-1\right)}{3}\)
4A = 4+4^2+....+4^51
3A = 4A - A = (4+4^2+....+4^51) - (1+4+4^2+....+4^50) = 4^51 - 1
=> A = (4^51-1)/3
k mk nha
a, C = 1 + 4 + 42 + 43 + 44 + 45 + 46
4C = 4 + 42 + 43 + 44 + 45 + 46 + 47
b, 4C - C = ( 4+42 + 43 + 44 +45 + 46 + 47 ) - ( 1 + 4 + 42 + 43 +44 +45 + 46 )
3C = 47 - 1
=> C = ( 47 - 1 ) : 3
nhớ k đấy nhé
\(A=1+4+4^2+4^3+...+4^{50}\)
=> \(4A=4+4^2+4^3+4^4+...+4^{51}\)
=> \(4A-A=\left(4+4^2+4^3+...+4^{51}\right)-\left(1+4+4^2+...+4^{50}\right)\)
=> \(3A=4^{51}-1\)
=> \(A=\frac{4^{51}-1}{3}\)
a) Ta có C = 1 + 4 + 42 + ... + 46
4C = 4( 1 + 4 + 42 + ... + 46 )
= 4 + 42 + 43 + ... + 47
b) Ta có C = 1 + 4 + 42 + ... + 46
4C = 4 + 42 + 43 + ... + 47
⇒ 4C - C = ( 4 + 42 + 43 + ... + 47 ) - ( 1 + 4 + 42 + ... + 46 )
⇒ 3C = 4 + 42 + 43 + ... + 47 - 1 - 4 - 42 - ... - 46
⇒ 3C = 47 - 1
⇒ C = \(\dfrac{4^7-1}{3}\) ( đpcm )
\(A=1+4+4^2+...+4^{2017}\)
=>\(4\cdot A=4+4^2+4^3+...+4^{2018}\)
=>\(4A-A=4+4^2+...+4^{2018}-1-4-4^2-...-4^{2017}\)
=>\(3A=4^{2018}-1\)
=>\(A=\dfrac{4^{2018}-1}{3}\)
\(2B-A=\dfrac{4^{2018}}{6}\cdot2-\dfrac{4^{2018}-1}{3}\)
\(=\dfrac{4^{2018}}{3}-\dfrac{4^{2018}-1}{3}=\dfrac{1}{3}\)
\(A=1-4+4^2-4^3+...+4^{98}-4^{99}+4^{100}\)
=>\(4A=4-4^2+4^3-4^4+...+4^{99}-4^{100}+4^{101}\)
=>\(4A+A=4-4^2+4^3-...+4^{99}-4^{100}+4^{101}+1-4+4^2-...+4^{98}-4^{99}+4^{100}\)
=>\(5A=4^{101}+1\)
=>\(A=\dfrac{4^{101}+1}{5}\)