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A=32019+1+3+32+33+...+32018
⇒A=1+3+32+...+32018+32019
⇒3A=3×(1+3+3^2+3^3+....+3^2019)
3A=3+3^2+3^3+....+3^2020
3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)
2A= 3^2020-1
⇒ A =( 3^2020-1):2
A=32019+1+3+32+33+...+32018
⇒A=1+3+32+...+32018+32019
⇒3A=3×(1+3+3^2+3^3+....+3^2019)
⇒3A=3+3^2+3^3+....+3^2020
⇒3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)
⇒2A= 3^2020-1
⇒ A =( 3^2020-1):2
\(A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)
\(=4\left(1+3^2+...+3^{2018}\right)⋮4\)
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
B = 1 + 32 + 34 + … + 32018
32.B = 32.( 1 + 32 + 34 + … + 32018)
9B = 32 + 34 + 36 + … + 32020
9B – B = (32 + 34 + 36 + … + 32020) – (1 + 32 + 34 + … + 32018)
8B = 32020 – 1
B = (32020 – 1) : 8.
Vậy B = (32020 – 1) : 8.
A = 1 + 3 + 32 + 33 + ... + 3100
3A = 3 + 32 + 33 +34+ .... + 3101
3A - A = (3 + 32 + 34 + ... + 3101) - (1 + 3 + 32 + 33 + ... + 3100)
2A = 3 + 32 + 34 + ... + 3101 - 1 - 3 - 32 - 33 - ... - 3100
2A = (3 - 3) + (32 - 32) + ... + (3100 - 3100) + (3101 - 1)
2A = 3101 - 1
A = \(\dfrac{3^{101}-1}{2}\)
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) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
\(3B=3+3^2+3^3+...+3^{2019}\\ 2B=3^{2019}-1\\ B=\dfrac{3^{2019}-1}{2}\)
\(9B=3^2+3^4+...+3^{2020}\)
\(\Leftrightarrow8B=3^{2018}-1\)
\(\Leftrightarrow B=\dfrac{3^{2018}-1}{8}\)
\(A=2\cdot\left(1+3+3^2+3^3+.....+3^{2018}\right)+1\)
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