Đặt

\(A=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+.......+n\left(n+1\right)\left(n+2\right)\)\(4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+3\cdot4\cdot5\cdot4+.......+n\left(n+1\right)\left(n+2\right)\cdot4\)\(4A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+........+n\left(n+1\right)\left(n+2\right)\left(n+3-n-1\right)\)\(4A=1\cdot2\cdot3\cdot4-0+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+....+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)\(4A=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(A=\dfrac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

Vậy \(A=\dfrac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

4N = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 2015.2016.2017.(2018-2014)

4N = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2015.2016.2017.2018 - 2014.2015.2016.2017

4N = (1.2.3.4 + 2.3.4.5 + 3.4.5.6 + ... + 2015.2016.2017.2018) - (0.1.2.3 + 1.2.3.4 + 2.3.4.5 + ... + 2014.2015.2016.2017)

4N = 2015.2016.2017.2018 - 0.1.2.3

4N = 2015.2016.2017.2018

N = 2015.2016.504.2018 (kq hơi to nên bn tự tính nhé)

4N = 1.2.3.4+ 2.3.4.4 + .... + 19.20.21.4

4N = 1.2.3.(4-0) + ...+ 19.20.21.(22-18)

4N = 1.2.3.4 - 0.1.2.3 + .... + 19.20.21.22-18.19.20.21

4N = 19.20.21.22

N = 19.5.21.22

Ez thôi mà :)

B1: S = 1 + 2 + 3 + .. .+ n 

=> S = ( n + 1 ) . n : 2 = aaa

=> S = ( n + 1 ) . n = 2aaa

Ta có: aaa = 111 . a = 37 . 3 . a

=> 2aaa = 37 . 6 . a 

Mà ( n + 1 ) . n là 2 số tự nhiên liên tiếp => 6a = 36 => a = 6

=> ( n + 1 ) . n = 37 . 36

=> n = 36

B2: Đề sai thì phải -_- T sửa lại 

(x + 2) + (4x + 4) + (7x + 6) + ... + (25x + 18) + (28x + 20) = 1560

<=> (x + 4x + 7x + ... + 25x + 28x) + (2 + 4 + 6 + ... + 18 + 20) = 1560

<=> 145x + 110 = 1560

<=> 145x = 1450

<=> x = 10

Bài 1:

uses crt;
var n,i,t1,t2:integer;
begin
clrscr;
write('Nhap n='); readln(n);
t1:=0;
for i:=1 to n do
t1:=t1+i*(i+1)*(i+2);
t2:=0;
for i:=1 to n do
begin
if i mod 2<>0 then t2:=t2+i*(i+1)*(i+2)
else t2:=t2-i*(i+1)*(i+2);
end;
writeln('T1=',t1);
writeln('T2=',t2);
readln;
end.

Bài 2:

uses crt;

var i,dem,n:integer;

begin

clrscr;

write('Nhap n='); readln(n);

dem:=0;

writeln('Cac uoc cua mot so ',n,' la: ');

for i:=1 to n do

if n mod i=0 then

begin

write(i:4);

dem:=dem+1;

end;

writeln;

writeln('So luong uoc cua ',n,' la: ',dem);

readln;

end.

4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]

4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4

4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]

4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)

4A = (n – 1).n(n + 1).(n + 2)

A = (n – 1).n(n + 1).(n + 2) : 4.

cau a thi sao ha ban ? 

ok thanks ban nhe

Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30

4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)

4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30

4A = 28.29.30.31 - 0.1.2.3

4A = 28.29.30.31

\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)

Theo cách tính trên ta dễ dàng tính được:

1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)

21320

Em nói thật em mới học lớp 6 Màu em đã phải làm bài này rồi thật đấu không phải đùa đâu

Ta có: 2/1.2.3=1/1.2-1/3.4

2/2.3.4=1/2.3-1/3.4

2/3.4.5=1/3.4-1/4.5

2/37.38.39=1/37.38-1/38.39

=>2A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+......+1/37.38-1/38.39

= 1/1.2-1/38.39=370/741

​=> A= 370/741

(x+2)+(4x+4)+(7x+6)+...+(25x+18)+(28x+20)=1560

(x+4x+7x+...+25x+28x)+(2+4+6+..+18+20)=1560

145x+110=1560

145x=1450

x=10

vậy........