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6 tháng 7 2016

Đặt \(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2016}}\)

\(=>3S=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2015}}\)

\(=>3S-S=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2015}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2016}}\right)\)

\(=>2S=1-\frac{1}{3^{2016}}=\frac{3^{2016}-1}{3^{2016}}=>S=\frac{3^{2016}-1}{3^{2016}}:2=\frac{3^{2016}-1}{2.3^{2016}}\)

Vậy...........
 

21 tháng 3 2017

bằng 15 hay sao ý

25 tháng 7 2020

Sửa để B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+....+\frac{1}{2016}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{1+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+....+\left(\frac{1}{2016}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2016}+\frac{2017}{2017}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)

20 tháng 3 2017

Đặt \(A=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.......+\frac{2}{2015}+\frac{1}{2016}\)

\(=\frac{2015}{2}+1+\frac{2014}{3}+1+...........+\frac{1}{2015}+1\)

\(=\frac{2017}{2}+\frac{2017}{3}+.........+\frac{2017}{2015}+\frac{2017}{2016}\)

\(=2017.\left(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2015}+\frac{1}{2016}\right)\)

Thay A vào biểu thức ta dc

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)

\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)

CÓ THỂ LÀ SAI NÊN BẠ THÔNG CẢM CHO MK

20 tháng 3 2017

sai rôi bạn ơi

24 tháng 9 2016

\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)

\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)

\(C=1+3:2+4:2+5:2+...+2017:2\)

\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)

\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)

\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)

\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)

\(C=2019.504=1017576\)

24 tháng 9 2016

sao lại chia 2

8 tháng 9 2016

\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)

\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)

\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{2015}{2017}\right)}\)

\(\Rightarrow A=2017\)

8 tháng 9 2016

\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)

\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)

\(A=\frac{4032}{\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{2}{2016.2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+\frac{2015}{2017}}\)

\(A=2017\)