tìm min p=\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}+\frac{101}{n+1}\)
Chứng minh với mọi số tự nhiên n khác 0 ta luôn có:
\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
Ta có :
\(\frac{1}{\sqrt{k}}=\frac{2}{2\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}\)
\(=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\)
\(=2\left(\sqrt{k+1}-\sqrt{k}\right)\)
Vậy : \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>2\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+....+2\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(=2\left(\sqrt{n+1}-1\right)\left(đpcm\right)\)
$\frac{1}{2\sqrt[3]{1}}$+$\frac{1}{3\sqrt[3]{2}}$+...+$\frac{1}{(n+1)\sqrt[3]{n}}$<3
Với n là số tự nhiên khác 0
Cho n là số tự nhiên khác 0. Tìm giá trị nhỏ nhất của
Q= \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+....+\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}+\frac{101}{n+1}\)
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\) \(=\sqrt{\frac{n^2\left(n+1\right)^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\frac{\left[n\left(n+1\right)+1\right]^2}{\left[n\left(n+1\right)\right]^2}}=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Do đó: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{n}-\frac{1}{n+1}+\frac{101}{n+1}\)
\(=n+1-\frac{1}{n+1}+\frac{101}{n+1}=n+1+\frac{100}{n+1}\ge2\sqrt{\left(n+1\right)\cdot\frac{100}{n+1}}=20\)
Dấu "=" \(\Leftrightarrow n+1=\frac{100}{n+1}\Leftrightarrow n=9\)
giả sử n là 1 số tự nhiên .chứng minh \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}<2\)
Xét \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\) = \(\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) < \(2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Vậy \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.....+\frac{1}{\left(n+1\right)\sqrt{n}}<2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) = \(2\left(1-\frac{1}{\sqrt{n+1}}\right)<2\) (đpcm)
a) Chứng tỏ rằng với số tưh nhiên n > 0 ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
b) Áp dụng kết quả trên hãy tính giá trị của biểu thức:
\(S=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2010^2}+\frac{1}{2011^2}}\)
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:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)
Tìm n là số N* biết
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+....._{ }+\sqrt{1+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}}=2001\frac{2001}{4006}\)
\(\sqrt{1+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}}=1-\frac{1}{n}+\frac{1}{n-1}\) dựa vào mà làm
\(\sqrt{1+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}}=\sqrt{\left(1-\frac{1}{n}\right)^2+\frac{2}{n}+\frac{1}{\left(n-1\right)^2}}\)
\(=\sqrt{\frac{\left(n-1\right)^2}{n^2}+\frac{2}{n}+\frac{1}{\left(n-1\right)^2}}\)
\(\sqrt{\left(\frac{n-1}{n}+\frac{1}{n-1}^2\right)}=1-\frac{1}{n}+\frac{1}{n-1}\)
Áp dụng đẳng thức vừa chứng minh vào phương trình trên ta được:
\(1-\frac{1}{3}+\frac{1}{2}+1-\frac{1}{4}+\frac{1}{3}+...+1-\frac{1}{n}+\frac{1}{n-1}=2001\frac{2001}{4006}\)
<=>(1+1+1+...+1(n-2 số 1))\(+\frac{1}{2}-\frac{1}{n}=2001\frac{2001}{4006}\)
<=>\(n-2+\frac{1}{2}-\frac{1}{n}=2001\frac{2001}{4006}\)
=>4006n.(n-2)+2003n-4006=8018007
=>4006n2-8012n+2003n-4006=8018007
=>4006n2-6009n-8022013=0
@@số to thế
đến bước chỉ còn ẩn n thì bấm máy tính nhé_ kết quả n= 2003
chứng minh \(\frac{1}{\sqrt{1.2}3}+\frac{1}{\sqrt{2.3}4}+....+\frac{1}{\sqrt{n\left(n+1\right)}\left(n+2\right)}\)<\(\frac{1}{\sqrt{2}}\)với mọi n là số tự nhiên
CMR : với mọi số tự nhiên n > 1, ta có :
a) \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)
b) \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
a) Ta có \(\frac{1}{n+k}>\frac{1}{2n}\)với k=1;2;...;n-1
=> \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}>\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}=\frac{n}{2n}=\frac{1}{2}\)
Mặt khác ta có \(\frac{1}{n+k}+\frac{1}{n\left(+\left(n+1-k\right)\right)}< \frac{3}{2n}\)
\(\Leftrightarrow3k^2+3nk+n+3k\forall k=1;2;...;n\)
Với k=1 ta có \(\frac{1}{n+1}+\frac{1}{n+n}< \frac{3}{2n}\)
Với k=2 ta có \(\frac{1}{n+2}+\frac{1}{n+\left(n-1\right)}< \frac{3}{2n}\)
..........................................
Với k=n ta có \(\frac{1}{n+n}+\frac{1}{n+1}< \frac{3}{2n}\)
Cộng từng vế của 2 BĐT trên ta được
\(2\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right)< \frac{3}{2n}+\frac{3}{2n}+....+\frac{3}{2n}=\frac{3n}{2n}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)(đpcm)
Không cần chứng minh \(\frac{1}{2}< \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\)
b) Ta có \(\frac{1}{\sqrt{k}}=\frac{1}{2\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)
Khi cho k=1,.....,n ta có:
\(1>2\left(\sqrt{2}-1\right)\)
\(\frac{1}{\sqrt{2}}>2\left(\sqrt{3}-\sqrt{2}\right)\)
................................
\(\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
Cộng từng vế BĐT trên ta có \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có đpcm
Bài 1: CMR
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+........+\frac{1}{\left(n+1\right)\sqrt{n}}>2,n\varepsilonℕ^∗\)
Bài 2: Cho S= \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{3\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
CMR S<\(\frac{1}{2}\)