Giải:

Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}>0_{\left(1\right)}.\)(Do S là phân số).

Ta lại có:

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}.\)

\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2014.2014}.\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2013.2014}.\)

\(< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}.\)

\(< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2013}-\dfrac{1}{2013}\right)-\dfrac{1}{2014}.\)

\(< 1+0+0+0+...+0-\dfrac{1}{2014}.\)

\(< 1-\dfrac{1}{2014}.\)

\(< \dfrac{2013}{2014}.\)

\(\Rightarrow S< 1_{\left(2\right)}.\) (do \(\dfrac{2013}{2014}< 1\)).

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) \(\Rightarrow\) \(0< S< 1.\)

\(\Rightarrow S\) không phải là số tự nhiên.

Vậy ta thu được \(đpcm.\)

~ Học tốt!!! ~

Ta thấy : \(S>0\) \(\left(1\right)\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

...............................

\(\dfrac{1}{2014^2}< \dfrac{1}{2013.2014}\)

\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..................+\dfrac{1}{2013.2014}\)

\(\Rightarrow S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2013}-\dfrac{1}{2014}\)

\(\Rightarrow S< 1-\dfrac{1}{2014}\)

\(\Rightarrow S< 1\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Rightarrow0< S< 1\Rightarrow S\) ko là số tự nhiên \(\rightarrowđpcm\)

bài 1

BÀI 3

uses crt;
var a: array[1..100] of integer;
i,n,max,s,j: integer;
begin
clrscr;
writeln(' nhap so phan tu cua day'); readln(n);
for i:=1 to n do
begin
writeln('a[',i,']'); readln(a[i]);
end;
max:=a[1];
s:=0;
for i:=1 to n do
begin
if max<a[i] then max:=a[i];
s:=s+a[i];
end;
writeln('so lon nhat trong day tren la:',max);
writeln('tong bang:',s);
writeln('so nguyen to trong mang la:');
j:=1;
for i:=1 to n do
if a[i]>1 then
begin
repeat
inc(j);
until (a[i] mod j=0);
if j>(a[i] div 2) then writeln(a[i]);
j:=1;
end;
readln
end.

#include <bits/stdc++.h>

using namespace std;

double s,a;

int i,n;

int main()

{

cin>>a;

s=0;

n=0;

while (s<=a) 

{

n=n+1;

s=s+1/(n*1.0);

}

cout<<n;

return 0;

}

1:

\(S=-\left(1-\dfrac{1}{10}+\dfrac{1}{10^2}-...-\dfrac{1}{10^{n-1}}\right)\)

\(=-\left[\left(-\dfrac{1}{10}\right)^0+\left(-\dfrac{1}{10}\right)^1+...+\left(-\dfrac{1}{10}\right)^{n-1}\right]\)

\(u_1=\left(-\dfrac{1}{10}\right)^0;q=-\dfrac{1}{10}\)

\(\left(-\dfrac{1}{10}\right)^0+\left(-\dfrac{1}{10}\right)^1+...+\left(-\dfrac{1}{10}\right)^{n-1}\)

\(=\dfrac{\left(-\dfrac{1}{10}\right)^0\left(1-\left(-\dfrac{1}{10}\right)^{n-1}\right)}{-\dfrac{1}{10}-1}\)

\(=\dfrac{1-\left(-\dfrac{1}{10}\right)^{n-1}}{-\dfrac{11}{10}}\)

=>\(S=\dfrac{1-\left(-\dfrac{1}{10}\right)^{n-1}}{\dfrac{11}{10}}\)

2:

\(S=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^{n-1}\)

\(u_1=1;q=\dfrac{1}{3}\)

\(S_{n-1}=\dfrac{1\cdot\left(1-\left(\dfrac{1}{3}\right)^{n-1}\right)}{1-\dfrac{1}{3}}\)

\(=\dfrac{3}{2}\left(1-\left(\dfrac{1}{3}\right)^{n-1}\right)\)

\(1,\) Ta có \(\left\{{}\begin{matrix}q=\dfrac{u_2}{u_1}=\dfrac{1}{10}:\left(-1\right)=-\dfrac{1}{10}\\u_1=-1\end{matrix}\right.\)

Vậy \(S=-1+\dfrac{1}{10}-\dfrac{1}{10^2}+...+\dfrac{\left(-1\right)^n}{10^{n-1}}=\dfrac{-1}{1-\left(-\dfrac{1}{10}\right)}=-\dfrac{10}{11}\)

\(2,\) Ta có \(\left\{{}\begin{matrix}q=\dfrac{u_2}{u_1}=\dfrac{1}{3}\\u_1=1\end{matrix}\right.\)

Vậy \(S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}\)

\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\\ S=\left(1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)\\ S=n-1-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)< n-1\)

Lại có \(\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{n^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)

\(\Rightarrow S>n-1-1=n-2\\ \Rightarrow n-2< S< n-1\\ \Rightarrow S\notin N\)

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.