Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
Đặt \(A=1+2+2^2+2^3+...+2^{2015}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)
\(\Rightarrow A=2^{2016}-1=-\left(1-2^{2016}\right)\) (Đặt dấu trừ ra trước thì đổi dấu)
Ta có: \(S=\frac{A}{1-2^{2016}}=\frac{-\left(1-2^{2016}\right)}{1-2^{2016}}=-1\)
Vậy S= -1
Có đc 1 GP ko nhỉ 
dễ thấy B=\(\frac{2015+2016}{2016+2017}\)<1
A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)=1-\(\frac{1}{2016}\)+1-\(\frac{1}{2017}\)=(1+1)-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))=2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))
vì (\(\frac{1}{2016}\)+\(\frac{1}{2017}\))<0,5+0,5=1 suy ra 2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))>1 mà b<1suy ra A>B
Ta thấy: B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)
A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
Mà\(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)
Suy ra: \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)>\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)=\(\frac{2015+2016}{2016+2017}\)
Hay A>B
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
Ta thấy: \(\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)
\(\left(\sqrt{a+b}\right)^2=a+b\)
Nếu: \(2\sqrt{ab}>0\left(a,b>0\right)\text{ thì: }\left(\sqrt{a}+\sqrt{b}\right)^2>\left(\sqrt{a+b}\right)^2\)
<=>\(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}+....+\frac{1}{\sqrt{2013}+\sqrt{2015}}\)
\(=\frac{1}{2}.\left(\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}+...+\frac{2}{\sqrt{2013}-\sqrt{2014}}\right)\)
\(=\frac{1}{2}.\left(-1+\sqrt{3}-\sqrt{3}+\sqrt{5}-...-\sqrt{2013}+\sqrt{2015}\right)\)
=\(\frac{\sqrt{2015}-1}{2}\)
Xét hiệu: B-A=\(\frac{\sqrt{2015}-1}{2}-\sqrt{481}=\frac{\sqrt{2015}-1}{2}-\frac{\sqrt{1924}}{2}=\frac{\sqrt{2015}-\left(\sqrt{1}+\sqrt{1924}\right)}{2}>\frac{\sqrt{2015}-\sqrt{1+1924}}{2}\)
\(=\frac{\sqrt{2015}-\sqrt{1925}}{2}>0\Rightarrow A>B\)
Ta có:
\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow2A=2.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}\right)=2.\frac{2015}{2017}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{4030}{2017}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{4030}{2017}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}=\frac{4030}{2017}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{4030}{2017}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{4030}{2017}\)
Bạn xem lại đề
A=-1/2*-2/3*-3/4*..*-2013/2014
A=-1*-2*-3*...*-2013/2*3*4*...*2014
A=-1/2014
ta có(-1)^2015=-1
B=-1/2015>-1/2014=A
nên A<B
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 \(S=\frac{8124480}{2017}\)
Sai vì ngoài học tập ra còn cần phải siêng năng chăm chỉ trong các lĩnh vực khác nửa như giúp đỡ mọi người ,tham gia các hoạt động rèn luyện