\(\frac{1}{\sqrt{1}+\sqrt{2}}+...\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\).Tìm n
Tìm n thỏa mãn \(A=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{\sqrt{2017}-1}{\sqrt{2017}}\)
Tìm \(\text{n}\inℕ\), biết :
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{\text{n}-1}+\sqrt{\text{n}}}=11\).
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
Bt: Tính
a) \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
b) C/m: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
c) C/m: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}
a, bạn chỉ cần lập công thức tông quát :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cái này bạn chỉ cần trục căn thức ở mẫu chưng minh xong áp dụng vào luôn là ra
a, kq : 4/5
b,\(1-\frac{1}{\sqrt{n+1}}\)
c,d chưa nghĩ ra
Bt: Tính
a) \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
b) C/m: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
c) C/m: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}<2\)
d) C/m: \(\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n}\)
ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{\left(n+1\right)n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{\left(n+1\right)n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
nên: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}=\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)\(=1-\frac{1}{5}=\frac{4}{5}\)
CMR
\(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+........+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)
\(\sqrt{1}+\sqrt{2}+.......+\sqrt{n}< n\sqrt{\frac{n+1}{2}}\)
\(\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+.........+\frac{\sqrt{25}-\sqrt{24}}{24+25}< \frac{2}{5}\)
Bài 1: CMR
Bài 2: CMR
Tìm số tự nhiên n nhỏ nhất sao cho\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\ge\)2014
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\ge2014\)
\(\Rightarrow\frac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+...+\frac{\sqrt{n}-\sqrt{n+1}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n}-\sqrt{n+1}\right)}\)
\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{n}-\sqrt{n+1}}{n-\left(n+1\right)}\)
\(=\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{n}-\sqrt{n+1}}{-1}\)
\(=\frac{1-\sqrt{n+1}}{-1}=\sqrt{n+1}-1\ge2014\)
\(\Leftrightarrow\sqrt{n+1}\ge2015\)
\(\Leftrightarrow n+1=2015^2=4060225\)
\(V~~n=4060224\)
a, Chứng minh
\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, Áp dụng
\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}......+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
a)\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b)\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\( S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
\(a,\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(n+1-n\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{\sqrt{n-1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
1) CMR \(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+\frac{1}{\sqrt{3.1997}}+...+\frac{1}{\sqrt{1999.1}}\ge1,999\)
2) CMR \(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{95\sqrt{94}+94\sqrt{95}}< 1\)
3) CMR \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
4) CMR \(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)