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1,\(A=\)\(1+2+2^2+2^3+...+2^{2015}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)
\(A=\)\(2^{2016}-1\)
~~~Hok tốt~~~
2,\(B=3^{11}+3^{12}+3^{13}+...+3^{101}\)
\(\Rightarrow3B=3^{12}+3^{13}+3^{14}+...+3^{102}\)
\(\Rightarrow3B-B=\left(3^{12}+3^{13}+3^{14}+...+3^{102}\right)-\left(3^{11}+3^{12}+3^{13}+...+3^{101}\right)\)
\(\Rightarrow2B=3^{102}-3^{11}\)
\(\Rightarrow B=\frac{3^{102}-3^{11}}{2}\)
~~~Hok tốt~~~
a ) \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+...+2^{2011}\right)-\left(2^0+2^1+...+2^{2010}\right)\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b ) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(\Rightarrow3B-B=\left(3+3^2...+3^{2011}\right)-\left(1+3+...+3^{2010}\right)\)
\(\Rightarrow2B=3^{2011}-1\)
\(\Rightarrow B=\frac{3^{2011}-1}{2}\)
Chúc bạn học tốt !!!
a)Ta có \(2A=2^2+2^3+...+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
Vậy \(A=2^{101}-2\)
b)
Ta có \(3A=3^2+3^3+...+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Vậy \(A=\frac{3^{101}-3}{2}\)
a, 13 + 23 = 1+8 = 9 = 32 = (-3)2
b, 13 + 23 + 33 = 1+8+27 = 36 = 62 = (-6)2
c, 13 + 23 + 33 + 43 = 1+8+27+64 = 100 = 102 = (-10)2
d, 13 + 23 + 33 + 43 = 1+8+27+64+125 = 225 = 152 = (-15)2