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Đặt A =\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}.n\)
=> A > \(\sqrt{n}\)
=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>\sqrt{n}\)(Đpcm)
Sau khi ib với Hoàng Nguyễn thì đề bài như sau
Tìm \(n\inℕ\)biết
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
ĐKXĐ: n > 1
Ta đi c/m bài toán tổng quát
\(\frac{1}{\sqrt{a-1}+\sqrt{a}}=\frac{\sqrt{a}-\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}\)
\(=\frac{\sqrt{a}-\sqrt{a-1}}{a-a+1}\)
\(=\sqrt{a}-\sqrt{a-1}\)
Áp dụng vào bài toán đc
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
\(\Leftrightarrow\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n}-\sqrt{n-1}=11\)
\(\Leftrightarrow\sqrt{n-1}-1=11\)
\(\Leftrightarrow\sqrt{n-1}=12\)
\(\Leftrightarrow n-1=144\)
\(\Leftrightarrow n=145\left(TmĐKXĐ\right)\)
Vậy n = 145
a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=>đpcm
b) Từ công thức trên ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)
không biết làm