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a) ta có: A = 3^0 + 3^1 + 3^2 + ...+ 3^100
=> 3A = 3^1 + 3^2 + 3^3 + ...+ 3^101
=> 3A-A = 3^101 - 3^0
2A = 3^101 - 1
\(A=\frac{3^{101}-1}{2}\)
b) D = 1 - 5 + 5^2 - 5^3 + ...+ 5^98 - 5^99
=> 5D = 5 - 5^2 + 5^3 - 5^4+...+ 5^99 - 5^100
=> 5D+D = -5^100 + 1
6D = -5^100 + 1
\(D=\frac{-5^{100}+1}{6}\)
\(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow A=2^2+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow 2A=2^3+2^3+2^4+2^5+...+2^{21}\)
\(\Rightarrow2A-A=2^{21}+2^3-2^2-2^2\)
\(\Rightarrow A=2^2\left(2^{19}+2-1-1\right)\)
\(\Rightarrow A=2^{19}.2^2=2^{21}\)
\(\text{A}=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow\text{A}=4+(2^2+2^3+...+2^{20})\)
\(\Rightarrow\text{A}-4=2^2+2^3+...+2^{20}\)
\(\Rightarrow2\text{(A}-4)=2^3+2^4+...+2^{21}\)
\(\text{A}-4=2(\text{A}-4)-(\text{A}-4)=(2^3+2^4+...+2^{21})-(2^2+2^3+...+2^{20})\)
\(\Rightarrow\text{A}-4=(2^3-2^3)+(2^4-2^4)+....+(2^{21}-2^2)\)
\(\Rightarrow\text{A}-4=2^{21}-4\)
\(\Rightarrow\text{A}=2^{21}-4+4\Rightarrow\text{A}=2^{21}\)
Chúc bạn hok tốt :>
Ta có:
\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}\)
\(=\frac{3^{10}.16}{3^9.16}=\frac{3.1}{1.1}=3\)
Vậy giá trị biểu thức A là 3
Mình làm ngắn gọn nhé.
\(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)
\(\Rightarrow A=2^{51}-1\)
\(B=1+3+...+3^{66}\)
\(3B=3+3^2+...+3^{67}\)
\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)
\(2B=3^{67}-1\)
\(B=\frac{3^{67}-1}{2}\)
A= 1+3+3^2+...+3^100
3A=3x( 1+3+3^2+...+3^100 )
3A-A=(3+3^2+...+3^101)-( 1+3+3^2+...+3^100 )
2A=3^101-1
A= \(\frac{3^{101}-1}{2}\)
B= 1+3^2+3^4+...+3^100
\(3^2B\)= 3^2x( 1+3^2+3^4+...+3^100)
9B-B= (3^2+3^4+..+3^102)-( 1+3^2+3^4+...+3^100 )
8B= 3^102-1
B=\(\frac{3^{102}-1}{8}\)