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A=1+3+32+33+...+32017+32018
=>3xA=(1+3+32+33+...+32017+32018)x3
=3+32+33+34+...+32018+32019
=>3xA -A=(3+32+33+34+...+32018+32019)-(1+3+32+33+...+32017+32018)
2xA=32019-1
=>A=(32019-1) :2
Vậy rút gọn A ta được:A= (32019-1):2
Chúc học giỏi^^
\(A=3^0+3^1+3^2+......+3^{2018}\)
\(3A=3.\left(3^0+3^1+3^2+.....+3^{2018}\right)\)
\(3A=3^1+3^2+3^3+........+3^{2019}\)
\(3A-A=\left(3^1+3^2+3^3+......+3^{2019}\right)-\left(3^0+3^1+3^2+.....+3^{2018}\right)\)
\(2A=3^{2019}-3^0\)
\(A=\left(3^{2019}-3^0\right):2\)
\(B=6^{10}+6^{11}+6^{12}+....+6^{2012}\)
\(6B=6.\left(6^{10}+6^{11}+6^{12}+.....+6^{2012}\right)\)
\(6B=6^{11}+6^{12}+6^{13}+.......+6^{2013}\)
\(6B-B=\left(6^{11}+6^{12}+6^{13}+......+6^{2013}\right)-\left(6^{10}+6^{11}+6^{12}+.......+6^{2012}\right)\)
\(5B=6^{2013}-6^{10}\)
\(B=\left(6^{2013}-6^{10}\right):5\)
Rút gọn :
A = 1+ 2 + 22 + 23 + ... + 259 + 260
B = 3 + 32 + 33 + 34 + ... + 32018 + 32019
Giúp mình với
A = 1 + 2 + 22 + 23 + ... + 259 + 260
2A = 2 + 22 + 23 + 24 + ... + 260 + 261
2A - A = 261 - 1
B = 3 + 32 + 33 + 34 + ... + 32018 + 32019
3B = 32 + 33 + 34 + 35 + ... + 32019 + 32020
3B - B = 32020 - 3
B = 32020−32
ta có
\(A=2^0+2^1+2^2+...+2^{60}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{61}\)
\(\Rightarrow2A-A=2^{61}-1\)
\(\Rightarrow A=2^{61}-1\)
tương tự với biểu thức B bạn lấy 3B - B còn 2B rồi chia cho 2 sẽ ra \(\frac{3^{2020}-3}{2}\)
Đặt \(D=1^2+2^2+3^2+...+2018^2\)
\(D=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+2018\left(2019-1\right)\)
\(D=1.2-1+2.3-2+3.4-3+...+2018.2019-2018\)
\(D=\left(1.2+2.3+...+2018.2019\right)-\left(1+2+3+...+2018\right)\)
Đặt \(A=1.2+2.3+...+2018.2019\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+2018.2019\left(2020-2017\right)\)
\(\Rightarrow3A=2018.2019.2010\Rightarrow A=\frac{2018.2019.2020}{3}\)
Đặt \(B=1+2+3+...+2018\)
\(B=\frac{\left(2018+1\right)\left(2018-1+1\right)}{2}=\frac{2019.2018}{2}\)
\(\Rightarrow D=A+B=\frac{2018.2019.2020}{3}+\frac{2019.2018}{2}\)
\(\Rightarrow D=\frac{2018.2019.2020.2+2019.2018.3}{6}\)
a, A = 1 + 3 + 3\(^{^2}\) + .... + 3\(^{100}\)
3A = 3 + 3\(^2\) + ..... + 3\(^{101}\)
Lấy 3A - A
\(\Rightarrow\) 2A = 3\(^{101}\) - 1
A = \(\frac{3^{101}-1}{2}\)
b, Áp dụng kiến thức câu a
A=1+2+22+23+...+22016
2A=2+22+23+24+...+22017
2A-A=(2+22+23+24+...+22017)-(1+2+22+23+...+22016)
A=22017-1
B=1+3+32+33+...+32014
3B=3+32+33+34+...+32015
3B-B=(3+32+33+34+...+32015)-(1+3+32+33+...+32014)
2B=32015-1
B=\(\frac{3^{2015}-1}{2}\)
\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
2; 3 tuong tu
1) A = 1 + 2 + 22 + 23 + .... + 22018
2A = 2 + 22 + 23 + 24 + ..... + 22019
2A - A = ( 2 + 22 + 23 + 24 + ..... + 22019 ) - ( 1 + 2 + 22 + 23 + .... + 22018 )
Vậy A = 22019 - 1
2) B = 1 + 3 + 32 + 33 + ..... + 32018
3A = 3 + 32 + 33 + ...... + 32019
3A - A = ( 3 + 32 + 33 + ...... + 32019 ) - ( 1 + 3 + 32 + 33 + ..... + 32018 )
2A = 32019 - 1
Vậy A = ( 32019 - 1 ) : 2
3) C = 1 + 4 + 42 + 43 + ...... + 42018
4A = 4 + 42 + 43 + ...... + 42019
4A - A = ( 4 + 42 + 43 + ...... + 42019 ) - ( 1 + 4 + 42 + 43 + ...... + 42018 )
3A = 42019 - 1
Vậy A = ( 42019 - 1 ) : 3
\(A=2\cdot\left(1+3+3^2+3^3+.....+3^{2018}\right)+1\)
Đặt: \(B=1+3+3^2+3^3+....+3^{2018}\)
\(\Rightarrow3B=3+3^2+3^3+3^4+....+3^{2019}\)
\(\Rightarrow3B-B=\left(3+3^2+3^3+3^4+....+3^{2019}\right)-\left(1+3+3^2+3^3+....+3^{2018}\right)\)
\(2B=3^{2019}-1\)
\(B=\frac{3^{2019}-1}{2}\)
\(\Rightarrow A=2\left(1+3+3^2+3^3+....+3^{2018}\right)+1\)
\(=2\cdot\frac{3^{2019}-1}{2}+1\)
\(=\frac{2.3^{2019}-1}{2}+1\)
\(=3^{2019}-1+1\)
\(=3^{2019}\)
A=2.(1+3+3^2+3^3+....+3^2018)+1
ĐẶT B= 1+3+3^2+3^3+....+3^2018
3B=3+3^2+3^3+3^4....+3^2019
=> 3B-B=(3+3^2+3^3+3^4+...+3^2019)-(1+3+3^2+3^3+...+3^2018)
2B=3^2019-1
B=\(\frac{3^{2019}-1}{2}\)
=> A=2.\(\frac{3^{2019}-1}{2}+1\)
A=3^2019-1+1
A=3^2019