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tui làm b nha do a không biết làm
A=5+32+33+...+32018
3A=15+33+34+...+32019
3A-A=(15+33+34+...+32019)-(5+32+33+...+32018)
2A=32019+15-(5+32)
2A=32019+15-14
2A=32019+1
2A-1=32019+1-1
2A-1=32019
vậy n = 2019
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)
=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101
=>1-1/(n+1)=100/101
=>1/(n+1)=1/101
=>n+1=101
=>n=100
A)\(M=1+3+3^2+...+3^9\)\(\Rightarrow3M=3+3^2+3^3+...+3^{10}\)\(\Rightarrow3M-M=\left(3+3^2+3^3+...+3^{10}\right)-\left(1+3+3^2+...+3^9\right)\)
\(\Rightarrow2M=3^{10}-1\)\(\Rightarrow2M+1=3^{10}\)\(\Rightarrow n=10\)
B) \(A=1+4^2+...+4^{99}\)\(\Rightarrow4A=4+4^3+4^4+...+4^{100}\)\(\Rightarrow4A-A=\left(4+4^3+4^4+...+4^{100}\right)-\left(1+4^2+...+4^{99}\right)\)
\(\Rightarrow3A=4^{100}+4-4^2-1\Rightarrow3A=4^{100}-13\Rightarrow3A+13=4^{100}\Rightarrow n=100\)
\(A=3+2^2+...+2^{99}\)
\(\Rightarrow A=1+2+2^2+...+2^{99}\)
\(\Rightarrow2A=2+2^2+...+2^{100}\)
\(\Rightarrow2A-A=2+2^2+...+2^{100}-1-2-...-2^{99}\)
\(\Rightarrow A=2^{100}-1\)
Thay A = 2100 - 1 vào A + 1 = 4^n , ta có:
\(2^{100}-1+1=4^n\)
\(\Rightarrow2^{100}=2^{2n}\)
\(\Rightarrow2n=100\Rightarrow n=50\)