(y+ 1/2) + (y + 1/4) + (y + 1/80 + ....... + (y + 1/1024)
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S
0
AM
1
NT
3
TN
1 tháng 8 2017
A=\(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)
A=\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
=>2A=2.(\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))
=>2A=\(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
=>2A-A=(\(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\))-(\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))
=>A=1-\(\dfrac{1}{2^{10}}\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)=1-\(\dfrac{1}{2^{10}}\)
1 tháng 8 2017
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\\ \Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\\ A=1-\dfrac{1}{2^{10}}\)
DT
0
23 tháng 10 2021
\(\left(x+1\right)^{2y}=1024\)
\(\Rightarrow\left[\left(x+1\right)^y\right]^2=32^2\)
\(\Rightarrow\left(x+1\right)^y=32\)
Do \(x,y\in Z\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=2\\y=5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)