Tính giá trị của biểu thức: 1.2.3 - 2.3.4 + 3.4.5 - 4.5.6 + … - 2416.2417.2418
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minh chi can ket qua thoi cung duoc ko can giai ra dau
ai lam dung minh kick
Đặt biểu thức trên = A
Xét : B = 1.2.3+2.3.4+....+n.(n+1).(n+2)
4B = 1.2.3.4+2.3.4.4+....+n.(n+1).(n+2).4
= 1.2.3.4+2.3.4.(5-1)+....+n.(n+1).(n+2).[(n+3)-(n-1)]
= 1.2.3.4+2.3.4.5-1.2.3.4+....+n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)
= n.(n+1).(n+2).(n+3)
=> B = n.(n+1).(n+2).(n+3)/4
=> A = 222315.222316.222317.222318/4
k mk nha
___Vương Tuấn Khải___
Ta có: B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070
Theo bài ra ta có:
B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
Ta nhân cả 2 vế với số 4 thì được phương trình như sau;
4*B = 4*(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
<=> 4*B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
<=> 4*B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
<=> 4*B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
<=> 4*B = 17.18.19.20
<=> 4*B = 116280
<=> B = 116280/4 = 29070
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)
\(M=1.2.3+2.3.4+3.4.5+...+47.48.49\)
\(4M=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+47.48.49.\left(50-46\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+47.48.49.50-46.47.48.49\)
\(=47.48.49.50\)
\(M=\frac{47.48.49.50}{4}=1381800\)
Bùi Lê Anh Khoa
Ta có: B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070
Ta có: B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070
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.
A=1.2.3+2.3.4+...+2416.2417.2418=(2416.2417.2418.2419):4
B=1.2.3-2.3.4+...+2415.2416.2417-2416.2417.2418
A+B=2(1.2.3+3.4.5+...+2415.2416.2417)=2C
Xét C=1.2.3+3.4.5+...+2415.2416.2417
=1.3(5-3)+3.5(7-3)+...+2415.2417.(2419-3)
=1.3.5+3.5.7+...+2415.2417.2419-3(1.3+3.5+...2415.2417)
=(1.3.5.7+3.5.7.(9-1)+...+2415.2417.2419.(2421-2413)):8-3.(1.3+1.3.5+3.5.(7-1)+...+2415.2417(2419-2413)):6
=2415.2417.2419.2421:8-3.(1.3+2415.2417.2419):6
=> B=2C-A
Ko chắc lắm
dê et ??? bieu thuc nay ko co lien quan đê gia tri cua thua so h cua 1 so