chứng minh : \(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^n}\)<1
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15 tháng 9 2021
\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)
\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)
AH
Akai Haruma
Giáo viên
13 tháng 4 2021
Lời giải:
\(u_{n+1}=\frac{n+2}{2^{n+2}}\left(\frac{2}{1}+...+\frac{2^{n+1}}{n+1}\right)=\frac{n+2}{2^{n+1}}\left(\frac{2^{n+1}}{n+1}u_n+\frac{2^{n+1}}{n+1}\right)=\frac{n+2}{2n+2}(u_n+1)\)
Ta chứng minh $u_n\geq 1(*)$ với mọi $n=1,2,...$
Thật vậy:
$u_1=1; u_2=\frac{3}{2}>1$. Giả sử $(*)$ đúng đến $n=k$
$u_{k+1}=\frac{k+2}{2k+2}(u_k+1)>\frac{2(k+2)}{2k+2}>1$
Do đó $u_n\geq 1$ với mọi $n=1,2,...$
Tiếp theo ta chứng minh $u_n< 1+\frac{4}{n}(**)$ với mọi $n=1,2,...$
Thật vậy:
$u_1=1< 1+\frac{4}{1}$
$u_2=\frac{3}{2}< 1+\frac{4}{2};....;u_4=\frac{5}{3}<1+\frac{4}{4}$
....
Giả sử $(**)$ đúng đến $n=k\geq 5$. Khi đó:
\(u_{k+1}=\frac{k+2}{2k+2}(u_k+1)<\frac{k+2}{2k+2}(2+\frac{4}{k})=\frac{(k+2)^2}{k(k+1)}\)
\(\frac{(k+2)^2}{k(k+1)}-(1+\frac{4}{k+1})=\frac{(k+2)^2-k(k+5)}{k(k+1)}=\frac{4-k}{k(k+1)}<0\) với mọi $k\geq 5$
$\Rightarrow u_{k+1}< 1+\frac{4}{k+1}$. Phép quy nạp hoàn tất.
Do đó $(**)$ đúng
Từ $(*); (**)\Rightarrow 1\leq u_n\leq 1+\frac{4}{n}$ với mọi $n=1,2,...$
Mà $\lim (1+\frac{4}{n})=1$ khi $n\to +\infty$ nên $\lim u_n=1$
TH
3
22 tháng 6 2017
Đặt :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+................+\dfrac{1}{2^n}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^{n-1}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{n-1}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^n}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^n}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...............+\dfrac{1}{2^n}< 1\rightarrowđpcm\)
JP
22 tháng 6 2017
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
................
\(\dfrac{1}{2^n}< \dfrac{1}{n.\left(n-1\right)}\)
\(\)- > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2^n}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)= \(1-\dfrac{1}{n}< 1\left(ĐPCM\right)\)
NM
14 tháng 6 2017
Áp dụng : \(\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n-1}}+...+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{2}}+1>2\left(\sqrt{n+1}-\sqrt{n}\right)+2\left(\sqrt{n}-\sqrt{n-1}\right)+...+2\left(\sqrt{4}-\sqrt{3}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+2\left(\sqrt{2}-1\right).\)
\(=2\left(\sqrt{n+1}-1\right).\)
14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
TD
28 tháng 2 2023
Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
AH
Akai Haruma
Giáo viên
28 tháng 2 2023
2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^n}\)
\(\Rightarrow2A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{n-1}}\)
\(A=2A-A=\dfrac{1}{2}-\dfrac{1}{2^n}< \dfrac{1}{2}< 1\)
Đặt A = \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^n}\)
2A = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{n-1}}\)
2A - A = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{n-1}}-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^n}\right)\)
A = \(\dfrac{1}{2}-\dfrac{1}{2^n}\)
Vì \(\dfrac{1}{2}-\dfrac{1}{2^n}< \dfrac{1}{2}\)
Mà \(\dfrac{1}{2}< 1\)
Nên \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...\dfrac{1}{2^n}< 1\)
Chúc học tốt!