( \(\frac{2048}{2049}\) + 12345 + 875 - 7 ) *\(\frac{1}{2016}\) * 0 = ???
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PN
6 tháng 5 2016
S= -(1/7^0 + 1/7^1+ 1/7^2 + 1/7^3 +...+ 1/7^2016)
Xét A = 1/7^0 + 1/7^1 + 1/7^2 + 1/7^3 +...+ 1/7^2016
=>7A= 7 + 1/7^0 + 1/7^1 + ...+ 1/7^2015
=> 6A = 7 - 1/7^2016
=> A = (7 - 1/7^2016)/6
=>S=-(7-1/7^2016)/6
T
4
11 tháng 8 2016
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-...-\frac{1}{2048}+\frac{1}{2048}\)
\(=1-\frac{1}{2048}\)
\(=\frac{2047}{2048}\)
k mk nha
PN
4
3 tháng 8 2015
\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(P=\frac{3\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)
\(\frac{1}{B}=3+3^2+3^3+...+3^{2015}+3^{2016}\)
\(\frac{3}{B}=3^2+3^3+3^4+...+3^{2016}+3^{2017}\)
\(\frac{3}{B}-\frac{1}{B}=\left(3^2+3^3+3^4+...+3^{2016}+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2015}+3^{2016}\right)\)
\(\frac{2}{B}=3^{2017}-3\)
\(B=\frac{2}{3^{2017}-3}\)
3 tháng 8 2015
P=\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
P=\(\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{3}+\frac{11}{7}+\frac{11}{3}}\)
P=\(\frac{\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{\frac{1}{11}.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
P=\(\frac{\frac{1}{3}}{\frac{1}{11}}=\frac{1}{3}:\frac{1}{11}=\frac{11}{3}\)
B=\(\frac{1}{3}+\frac{1}{^{3^2}}+\frac{1}{3^3}+................+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)
B=\(\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2015}+\left(\frac{1}{3}\right)^{2016}\)
2B=\(\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^4+...+\left(\frac{1}{3}\right)^{2016}+\left(\frac{1}{3}\right)^{2017}\)
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B=\(\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2015}+\left(\frac{1}{3}\right)^{2016}\)
B=\(\left(\frac{1}{3}\right)^1-\left(\frac{1}{3}\right)^{2017}\)
(2048/2049+12345+875-7)x1/2016x0
Ta biết mọi số nhân với 0 đều bằng 0
=>(2048/2049+12345+875-7)x1/2016x0=0
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