Câu 6:

uses crt;

var n,i:integer;

begin

clrscr;

readln(n);

for i:=1 to n do 

if n mod i=0 then write(i:4);

readln;

end.

5:

uses crt;

var n,i,dem:integer;

begin

clrscr;

readln(n);

dem:=0;

for i:=0 to n do

if i mod 2=1 then 

begin

write(i:4);

dem:=dem+1;

end;

writeln;

writeln(dem);

readln;

end.

uses crt;

var s:real;

i,n:integer;

begin

clrscr;

readln(n);

s:=0;

for i:=1 to n do 

  s:=s+(n*(n+1))/((n+2)*(n+3));

writeln(s:4:2);

readln;

end.

Bạn hãy tôn trọng người khác bằng cách chọn đúng tên lớp         

Nhầm ! Tại ghi lớp 5 quen ùi !

\(S=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(S=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)

\(T=\dfrac{3}{1x2}+\dfrac{3}{2x3}+\dfrac{3}{3x4}+\dfrac{3}{4x5}+...\dfrac{3}{nx\left(n+1\right)}\)

\(T=3x\left[\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\right]\)

\(T=3x\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\right]\)

\(T=3x\left(1-\dfrac{1}{n+1}\right)=\dfrac{3xn}{n+1}\)

A=(1/1.2.3-1/2.3.4)+(1/2.3.4-1/3.4.5)+..............+(1/n(n+1)(n+2)-1/(n+1)(n+2)(n+3))

A=1/1.2.3-1/(n+1)(n+2)(n+3)

A=1/18-1/(n+1)(n+2)(n+3)

đúng nhé

Đặt C= 1.2+2.3+3.4+...+n.(n+1)

3C=1.2.3+2.3.3+3.4.3+...+n.(n+1).3

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+....+n.(n+1)+[(n+2)-(n-1)]

3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n.(n+1).(n+2)-(n-1).n.(n+1)

3C=n.(n+1).(n+2)

C=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Câu 2: 

#include <bits/stdc++.h>

using namespace std;

double p1,p2;

int i,n;

int main()

{

cin>>n;

p1=1;

p2=1;

for (i=1; i<=n; i++)

{

if (i%2==0) p2=p2*(i*1.0);

else p1=p1*(i*1.0);

}

cout<<fixed<<setprecision(2)<<p1<<endl;

cout<<fixed<<setprecision(2)<<p2;

return 0;

}

1/

\(N=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)=\)

\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)

Đặt 

\(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)

Đặt

\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)

\(\Rightarrow N=A-B\)

2/

Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được

\(A=1^2+2^2+3^2+...+100^2\) 

Tính như câu 1

3/ Làm như bài 4

4/

\(S=1^2+3^2+5^2+...+99^2=\)

\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)

\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)

Đặt

\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\) 

Đặt

\(A=1.3+3.5+5.7+...+99.101\)

\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)

\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)

\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)

\(\Rightarrow S=A-2B\)

Bài 1:

\(N=1^2+2^2+3^3+...+99^2\)

\(N=1.1+2.2+3.3+...+99.99\)

\(N=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)

\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)

\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)

+) Tính \(A=1.2+2.3+3.4+...+99.100\)

Ta có:

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

+) Tính \(B=1+2+3+...+99\)

\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)

\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)

\(\Rightarrow N=A-B=333300-4950=328350\)

\(\Rightarrow N=328350\)

xin loi mik danh nham nhe bai do la 10000 nhe