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chứng minh bài toán theo cách quy nạp toán học.
Với n=2 suy ra:\(\frac{1}{3}+\frac{1}{4}>\frac{13}{14}\left(TM\right)\)
Giả sử bài toán trên đúng với mọi n=k,ta cần chứng minh nó đúng với n=k+1,tức là:
\(S_k=\frac{1}{k+2}+\frac{1}{k+3}+\frac{1}{k+4}+....+\frac{1}{2\left(k+1\right)}>\frac{13}{14}\)
Thật vậy:
\(\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2\left(k+1\right)}\)
\(=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)
\(=S_k+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\)
\(>\frac{13}{14}+\frac{2k+2}{2\left(k+1\right)\left(2k+1\right)}+\frac{2k+1}{2\left(k+1\right)\left(2k+1\right)}-\frac{2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{14}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
để dễ hiểu,,mik xin viết thêm nha(không phải để kiếm điểm,có người nhờ nên mới thế này:))
\(\frac{13}{14}+\frac{2\left(k+1\right)+2k+1-2\left(2k+1\right)}{2\left(k+1\right)\left(2k+1\right)}\)
\(=\frac{13}{14}+\frac{1}{2\left(k+1\right)\left(2k+1\right)}>\frac{13}{14}\left(k>1\right)\)
\(\Rightarrow S_{k+1}>\frac{13}{14}\)
\(\Rightarrow S_k>\frac{13}{14}\)
Phép chứng minh hoàn tất_._
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)
\(\Rightarrow n+1=50\)
\(\Rightarrow n=49\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)
\(\Rightarrow2n+1=51\)
\(\Rightarrow2n=50\)
\(\Rightarrow n=25\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=....=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_2}{a_3}\right)^n=....=\left(\frac{a_n}{a_{n+1}}\right)^n=\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)(1)
Ta có: \(\left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}....\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}....\frac{a_n}{a_{n+1}}=\frac{a_1}{a_{n+1}}\)(2)
Từ (1), (2) \(\Rightarrow\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)(đpcm)
\(\text{Áp dụng tính chất của dãy tỉ số bằng nhau có:}\)
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_2}{a_3}\right)^n=...=\left(\frac{a_n}{a_{n+1}}\right)^n\)\(=\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)
Mà\( \left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}\cdot\frac{a_1}{a_2}\cdot...\cdot\frac{a_1}{a_2}\)\(=\frac{a_1}{a_2}\cdot\frac{a_2}{a_3}\cdot...\cdot\frac{a_n}{a_{n+1}}\)\(=\frac{a_1}{a_{n-1}}\)
\(\Rightarrow\)\(\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)\(=\frac{a_1}{a_{n-1}}\)
a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)
b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)
Bạn tham khảo cách làm ở đây: https://olm.vn/hoi-dap/question/528628.html
TA CÓ :\(\frac{1}{n+1}>\frac{1}{2n},\frac{1}{n+2}>\frac{1}{2n},....\)\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+...+\frac{1}{2n}\)(n số)
=\(\frac{n}{2n}=\frac{1}{2}\left(đcpm\right)\)