Cho P = (1-\(\dfrac{1}{1+2}\)) (1- \(\dfrac{1}{1+2+3}\)) … ( 1 -\(\dfrac{1}{1+2+...+n}\) ) . Tìm các số tự nhiên n \(\text{≥}\) 2 để \(\dfrac{1}{P}\) là số nguyên.
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Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)
=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101
=>1-1/(n+1)=100/101
=>1/(n+1)=1/101
=>n+1=101
=>n=100
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)
Ta có \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};...;\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)
Do đó \(a< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}=1+\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
\(=1+1-\dfrac{1}{n}=1-\dfrac{1}{n}< 2\) . Suy ra \(1< a< 2\)
Vậy \(a\) khôg phải số tự nhiên
Ta có: `1 < 1 + 1/2^2 + ... + 1/n^2`
`1/(2.2) < 1/(1.2)`
`1/(3.3) < 1/(2.3)`
`...`
`1/(n^2) < 1/(n-1(n))`
`=> 1/2^2 + ... + 1/n^2 < 1/(1.2) + ... + 1/(n-1(n)) = 1/1 - 1/n < 1`.
`=> a < 1 + 1 = 2`.
`=> 1 < a < 2`.
`=>` Đây không là số tự nhiên.
a) \(\dfrac{n+2}{3}\) là số tự nhiên khi
\(n+2⋮3\)
\(\Rightarrow n+2\in\left\{1;3\right\}\)
\(\Rightarrow n\in\left\{-1;1\right\}\left(n\in Z\right)\)
b) \(\dfrac{7}{n-1}\) là số tự nhiên khi
\(7⋮n-1\)
\(\Rightarrow7n-7\left(n-1\right)⋮n-1\)
\(\Rightarrow7n-7n+7⋮n-1\)
\(\Rightarrow7⋮n-1\)
\(\Rightarrow n-1\in\left\{1;7\right\}\Rightarrow\Rightarrow n\in\left\{2;8\right\}\left(n\in Z\right)\)
c) \(\dfrac{n+1}{n-1}\) là sô tự nhiên khi
\(n+1⋮n-1\)
\(\Rightarrow n+1-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+1-n+1⋮n-1\)
\(\Rightarrow2⋮n-1\)
\(\Rightarrow n-1\in\left\{1;2\right\}\Rightarrow n\in\left\{2;3\right\}\left(n\in Z\right)\)
(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì 2n⋮n-2
2n-4+4⋮n-2
2n-4⋮n-2⇒4⋮n-2
n-2∈Ư(4)⇒Ư(4)={1;-1;2;-2;4;-4}
n∈{3;1;4;0;6;-2}
(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{48.50}\)
=\(\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+...+\dfrac{2}{48.50}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)
=\(\dfrac{3}{2}.\dfrac{2}{25}\)
=\(\dfrac{3}{25}\)
Giải:
(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì \(2n⋮n-2\)
\(2n⋮n-2\)
\(\Rightarrow2n-4+4⋮n-2\)
\(\Rightarrow4⋮n-2\)
\(\Rightarrow n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
n-2 | -4 | -2 | -1 | 1 | 2 | 4 |
n | -2 | 0 | 1 | 3 | 4 | 6 |
Kết luận | loại | t/m | t/m | t/m | t/m | t/m |
Vậy \(n\in\left\{0;1;3;4;6\right\}\)
(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+\dfrac{3}{14.16}+...+\dfrac{3}{48.50}\)
\(=\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{48.50}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)
\(=\dfrac{3}{2}.\dfrac{2}{25}\)
\(=\dfrac{3}{25}\)
Chúc bạn học tốt!
a/ \(\lim\limits\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}=\lim\limits\dfrac{\dfrac{\left(\dfrac{1}{3}\right)^{n+1}-1}{\dfrac{1}{3}-1}}{\dfrac{\left(\dfrac{1}{2}\right)^{n+1}-1}{\dfrac{1}{2}-1}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}=3\)
b/ \(\lim\limits\left(n^3+n\sqrt{n}-5\right)=+\infty-5=+\infty\)
Lời giải:
Xét thừa số tổng quát:
\(1-\frac{1}{1+2+...+n}=\frac{(1+2+...+n)-1}{1+2+...+n}=\frac{\frac{n(n+1)}{2}-1}{\frac{n(n+1)}{2}}=\frac{n(n+1)-2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}\)
Thay $n=2,3,....,$ ta được:
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{(n-1)(n+2)}{n(n+1)}\)
\(=\frac{[1.2.3....(n-1)][4.5.6..(n+2)]}{(2.3.4..n)[3.4.5...(n+1)]}\)
\(=\frac{1}{n}.\frac{n+2}{3}=\frac{n+2}{3n}\)
\(\frac{1}{P}=\frac{3n}{n+2}\in\mathbb{Z}\) khi mà $3n\vdots n+2$
$\Leftrightarrow 3(n+2)-6\vdots n+2$
$\Leftrightarrow 6\vdots n+2$
$\Rightarrow n+2\in\left\{6\right\}$ (do $n+2\geq 4$ với mọi $n\geq 2$)
$\Rightarrow n=4$