tính tổng 1*2+2-3+3*4+....+n*n+1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
\(T=\dfrac{3}{1x2}+\dfrac{3}{2x3}+\dfrac{3}{3x4}+\dfrac{3}{4x5}+...\dfrac{3}{nx\left(n+1\right)}\)
\(T=3x\left[\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\right]\)
\(T=3x\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\right]\)
\(T=3x\left(1-\dfrac{1}{n+1}\right)=\dfrac{3xn}{n+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] : 3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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;
}
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
var n, i: integer; S: real;
begin
read(n);
while i <= n do
begin
S:=S+i*(i+1);
i:=i+1;
end;
write(S);
end.
b,
var n,i: integer; S: real;
begin
read(n);
while i <= n do
begin
S:=S+2*i;
i:=i+1;
end;
write(S);
end.
c,
var n,i: integer; S: real;
begin
read(n);
while i <=n do
begin
S:=S+(2*i+1);
i:=i+1;
end;
write(S);
end.
T CHỊU
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=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)