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\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\\ S=\left(1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)\\ S=n-1-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)< n-1\)
Lại có \(\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{n^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)
\(\Rightarrow S>n-1-1=n-2\\ \Rightarrow n-2< S< n-1\\ \Rightarrow S\notin N\)
Lời giải:
$n=1$ thì $S=0$ nguyên nhé bạn. Phải là $n>1$
\(S=1-\frac{1}{1^2}+1-\frac{1}{2^2}+1-\frac{1}{3^2}+...+1-\frac{1}{n^2}\)
\(=n-\underbrace{\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)}_{M}\)
Để cm $S$ không nguyên ta cần chứng minh $M$ không nguyên. Thật vậy
\(M> 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(M>1+\frac{1}{2}-\frac{1}{n+1}>1\) với mọi $n>1$
Mặt khác:
\(M< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
\(M< 1+1-\frac{1}{n}< 2\)
Vậy $1< M< 2$ nên $M$ không nguyên. Kéo theo $S$ không nguyên.
Câu hỏi của Nguyễn Thái Hà - Toán lớp 6 - Học toán với OnlineMath
Bạn tham khảo nhé!
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.
\(S=\dfrac{1}{2018}\left(1+\dfrac{1}{1}+1+\dfrac{1}{2}+1+\dfrac{1}{3}+...+1+\dfrac{1}{2018}\right)\)
\(S=\dfrac{1}{2018}\left(2018+\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
\(S=1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
Do \(\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2018}\right)>0\Rightarrow S>1\) (1)
Lại có:
\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}< \dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+...+\dfrac{1}{1}=2018\)
\(\Rightarrow1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)< 1+\dfrac{1}{2018}.2018=2\)
\(\Rightarrow S< 2\) (2)
Từ (1), (2) \(\Rightarrow1< S< 2\)
\(\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải là số tự nhiên
\(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=2-\dfrac{1}{n}< 2\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2\left(đpcm\right)\)
Vậy...
Để \(x=\dfrac{\sqrt{n-1}}{2}\) là số nguyên thì \(\sqrt{n-1}⋮2\)
=>\(n-1=\left(2k\right)^2=4k^2\)(k\(\in\)Z) và n>=1
=>\(n=4k^2+1\)
n<30
=>\(4k^2+1< 30\)
=>\(4k^2< 29\)
=>\(k^2< \dfrac{29}{4}\)
mà k nguyên
nên \(k^2\in\left\{0;1;4\right\}\)
\(n=4k^2+1\)
=>\(n\in\left\{1;5;17\right\}\)
Theo bài ra, ta có:
\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{16}\)
\(\Rightarrow S=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}\right)+\left(\dfrac{1}{15}+\dfrac{1}{16}\right)\)
\(\Rightarrow S< \left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{6}.3+\dfrac{1}{9}.3+\dfrac{1}{12}.3+\dfrac{1}{15}.3\)
\(\Rightarrow S< \left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(\Rightarrow S< 2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)\)
\(\Rightarrow S< 2\left[\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)\right]\)
\(\Rightarrow S< 2\left(\dfrac{2}{2}+\dfrac{2}{4}\right)\)
\(\Rightarrow S< 2\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)
\(\Rightarrow S< 2.\dfrac{3}{2}\)
\(\Rightarrow S< 3\left(1\right)\)
Lại có: \(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{16}\)
\(\Rightarrow S=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\dfrac{1}{8}.4+\dfrac{1}{12}.4+\dfrac{1}{16}.4\)
\(\Rightarrow S>\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)\)
\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}\right)\)
\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{2}{4}\right)\)
\(\Rightarrow S>2\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(\Rightarrow S>2\)
Từ (1) và (2) suy ra \(2< S< 3\)
⇒ S không phải 1 số nguyên
Vậy...
\(S=\dfrac{1^2}{1}-\dfrac{1}{1}+\dfrac{2^2}{2^2}-\dfrac{1}{2^2}+...+\dfrac{n^2}{n^2}-\dfrac{1}{n^2}\)
\(S=1-\dfrac{1}{1}+1-\dfrac{1}{2^2}+...+1-\dfrac{1}{n^2}\)
\(S=n-\left(\dfrac{1}{1}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)=n-A\)
Xét \(A=\dfrac{1}{1}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}=1+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}>1\) (1)
\(A=\dfrac{1}{1}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1}+\dfrac{1}{1.2}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow A< \dfrac{1}{1}+\dfrac{1}{1}-\dfrac{1}{2}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=2-\dfrac{1}{n}< 2\) (2)
Từ (1),(2) \(\Rightarrow1< A< 2\Rightarrow A\) nằm giữa 2 số nguyên liên tiếp nên A không phải là số nguyên.
Mà \(S=n-A\), do \(n\) nguyên, \(A\) không nguyên \(\Rightarrow S\) không nguyên