(12015-1012015) .( 22015-1002015) . (32015-992015)......(1012015-12015)=
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\(3^{x+3}\cdot2=5^3+37\cdot1^{2015}\\\Rightarrow3^{x+3}\cdot2=125+37\\\Rightarrow3^{x+3}\cdot2=162\\\Rightarrow3^{x+3}=162:2\\\Rightarrow3^{x+3}=81\\\Rightarrow3^{x+3}=3^4\\\Rightarrow x+3=4\\\Rightarrow x=4-3\\\Rightarrow x=1\)
\(3^{x+3}.2=5^3+37.1^{2015}\\ 3^{x+3}.2=125+37.1=125+37=162\\ 3^{x+3}=\dfrac{162}{2}=81=3^4\\ Nên:x+3=4\\ Vậy:x=4-3=1\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\)
\(2A=\frac{1}{3}-\frac{1}{57}\)
\(2A=\frac{6}{19}\)
\(A=\frac{3}{19}\)
Ta có:
A=1/15+1/35+1/63+1/99+...+1/2015+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/45.47+1/47.49
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/45-1/47+1/47-1/49
=1/3-1/49
=49/147-3/147
=47/147
Chị ơi, nói hay lắm bói cái đó ở đâu vậy, bạch dương và bảo bình ý?
Theo nhị thức Newton ta có:
( 3 + x ) 2015 = C 2015 0 . 3 2015 + C 2015 1 .3 2014 . x + C 2015 2 .3 2013 . x 2 + .... + C 2015 2014 .3. x 2014 + C 2015 2015 . x 2015
Thay x = -1 ta được:
( 3 − 1 ) 2015 = C 2015 0 . 3 2015 − C 2015 1 .3 2014 + C 2015 2 .3 2013 − .... + C 2015 2014 .3 − C 2015 2015
Suy ra, S = 2 2015
Ta chọn đáp án A
\(A=3+3^2+3^3+...+3^{2015}\)
\(\Rightarrow3A=3^2+3^3+...+3^{2015}+3^{2016}\)
\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)
\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2016}-3\right)\)
\(\Rightarrow2A=3^{2016}-3\)
\(\Rightarrow A=\dfrac{3^{2016}-3}{2}\)
Ta có: \(2A+3=3^n\)
\(\Rightarrow2\cdot\dfrac{3^{2016}-3}{2}+3=3^n\)
\(\Rightarrow3^{2016}-3+3=3^n\)
\(\Rightarrow3^{2016}=3^n\)
\(\Rightarrow n=2016\)