so sánh:
\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\)với \(\frac{1}{2}\)
Những câu hỏi liên quan
So sánh:
\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\)với \(\frac{1}{2}\)
so sánh:
\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\)với \(\frac{1}{2}\)
so sánh:
\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\)với \(\frac{1}{2}\)
So sánh:
A=\(\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\)với\(\frac{1}{2}\)
So sánh:
A=\(\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\)với \(\frac{1}{2}\)
so sánh:
\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2015}{4^{2015}}\) với \(\frac{1}{2}\)
Nhới giải chi tiết giùm mình nhé
Cho M=\(\frac{\sqrt{2}-\sqrt{1}}{1+1}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với 1/2
Cho \(M=\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\). Hãy so sánh M với \(\frac{1}{2}\)
so sánh biểu thức P với \(\frac{1}{2}\)biết
\(P=\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{2017}{2015!+2016!+2017!}\)(với n!=1.2.3...n)
\(P=\frac{3}{1!\left(1+2\right)+3!}+\frac{4}{2!\left(1+3\right)+4!}+...+\frac{2017}{2015!\left(1+2016\right)+2017!}\)
\(P=\frac{3}{3\left(1!+2!\right)}+\frac{4}{4\left(2!+3!\right)}+...+\frac{2017}{2017\left(2015!+2016!\right)}\)
\(P=\frac{1}{1!+2!}+\frac{1}{2!+3!}+...+\frac{1}{2015!+2016!}\)
Ta có \(a!>\sqrt{a}\)\(\left(a\inℕ;a>1\right)\) do đó :
\(P>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\)
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2016}\)
\(-\sqrt{2015}=\sqrt{2016}-1=\frac{1}{2}+\left(\sqrt{2016}-\frac{3}{2}\right)=\frac{1}{2}+\left(\sqrt{2016}-\sqrt{\frac{9}{4}}\right)>\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Chúc bạn học tốt ~
PS : tự nghĩ bừa thui nhé :))
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