tính tổng A = \(1+2+2^3+...+2^{2015}\)
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HT
0
PT
3
15 tháng 2 2017
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)16
2A=\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2017}\)
2A-A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)-\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)
A=\(\frac{1}{2017}-\frac{1}{2}\)
7 tháng 4 2016
Ta có :
\(S=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+..+2016}\)
\(=2015.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}\right)\)
\(=2015.\left(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\right)\)
\(=2015.\left(\frac{2}{2}+\frac{2}{2.\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.\left(2+1\right)}+\frac{1}{3.\left(3+1\right)}+...+\frac{1}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2017}\right)\)
\(=2015.2.\left(1-\frac{1}{2017}\right)\)
\(=2015.2.\frac{2016}{2017}\)
=\(\frac{2015.2.2016}{2017}\)
=\(\frac{8124480}{2017}\)
Vậy \(S=\frac{8124480}{2017}\)
NT
2
26 tháng 1 2018
Có : 2A = 1 + 1/2 + 1/2^2 +.....+ 1/2^2015
A = 2A - A = (1+1/2+1/2^2+.....+1/2^2015)-(1/2+1/2^2+.....+1/2^2016)
= 1 - 1/2^2016
Tk mk nha
26 tháng 1 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2015}}-\frac{1}{2^{2016}}\)
\(A=1-\frac{1}{2^{2016}}\)
S
14 tháng 2 2020
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
DT
0
Đặt \(A=1+2+2^3+...+2^{2015}\)
\(\Rightarrow2^2A=2^2+2^3+2^5+...+2^{2017}\)
\(\Rightarrow2^2A-A=3A=\left(2^{2017}+2^2\right)-\left(1+2\right)=2^{2017}+4-3=2^{2017}+1\)
Do đó \(A=\frac{2^{2017}+1}{3}\)