Rut gon tong sau
C=1+2^2+2^4+....+2^2008
K
Khách
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OL
4
28 tháng 7 2016
\(C=1+2^2+2^4+...+2^{2008}\)
\(\Rightarrow4C=2^2+2^4+...+2^{2010}\)
\(\Rightarrow4C-C=\left(2^2+2^4+...+2^{2010}\right)-\left(1+2^2+2^4+...+2^{2008}\right)\)
\(3C=2^{2010}-1\)
\(C=\frac{2^{2010}-1}{3}\)
28 tháng 7 2016
\(C=1+2^2+2^4+....+2^{2008}\)
\(\Rightarrow4C=2^2+2^4+.....+2^{2010}\)
\(\Rightarrow3C=4C-C=\left(2^2+2^4+...+2^{2010}\right)-\left(1+2^2+.....+2^{2008}\right)\)
\(\Rightarrow3C=2^{2010}-1\)
\(\Rightarrow C=\frac{2^{2010}-1}{3}\)
T
1
5 tháng 10 2019
\(3\left(2^2+1\right).\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right).\left(2^8+1\right)\left(2^{16}+1\right)....\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right).\left(2^{64}+1\right)+1\)
\(=2^{64}-1+1=2^{64}\)
Vậy : \(3\left(2^2+1\right).\left(2^4+1\right)...\left(2^{64}+1\right)+1=2^{64}\)
DK
1
27 tháng 7 2017
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}:\frac{2\left(x+1\right)^2}{4\left(x-1\right)^2}=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}.\frac{4\left(x-1\right)^2}{2\left(x+1\right)^2}=2\)
14 tháng 12 2017
Ta có:
\(\frac{2}{5}=\frac{2x7}{5x7}=\frac{14}{35}\)
mà 14+35=49 và rút gọn \(\frac{14}{35}=\frac{2}{5}\)nên phân số cần tìm là \(\frac{14}{35}\)
LV
4
N
26 tháng 10 2016
=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=...=2^32-1
TD
0
VN
0
NT
0
\(C=1+2^2+2^4+...+2^{2008}\)
\(4C=2^2+2^4+2^6+...+2^{2010}\)
\(4C-C=\left(2^2+2^4+2^6+...+2^{2010}\right)-\left(1+2^2+2^4+...+2^{2008}\right)\)
\(3C=2^{2010}-1\)
\(C=\frac{2^{2010}-1}{3}\)
\(C=1+2^2+2^4+...+2^{2008}\)
\(\Rightarrow4C=2^2+2^4+...+2^{2010}\)
\(\Rightarrow4C-C=\left(2^2+2^4+...+2^{2010}\right)-\left(1+2^2+2^4+...+2^{2008}\right)\)
\(\Rightarrow3C=2^{2010}-1\)
\(\Rightarrow C=\frac{2^{2010}-1}{3}\)