Chứng minh rằngA = 3 + 32+ 33+ ... + 32019+ 32020chia hết cho 10.
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\(A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)
\(=4\left(1+3^2+...+3^{2018}\right)⋮4\)
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
\(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+...+\frac{1}{60}\right)>\frac{1}{45}.15+\frac{1}{60}.15=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
=>đpcm
l-i-k-e cho mình nha
\(16^{10}+32^7=\left(2^4\right)^{10}+\left(2^5\right)^7=2^{40}+2^{35}=2^{35}.2^5+3^{35}=2^{35}.\left(2^5+1\right)=2^{35}.33\)
chia hết cho 33
tick nhé
Ghi lại đề: \(A=3+3^2+...+3^{2020}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2017}+3^{2018}+3^{2019}+3^{2020}\right)\\ A=3\left(1+3+3^2+3^3\right)+...+3^{2017}\left(1+3+3^2+3^3\right)\\ A=\left(1+3+3^2+3^3\right)\left(3+...+3^{2017}\right)\\ A=40\left(3+...+3^{2017}\right)⋮10\left(40⋮10\right)\)