Rút gọn phân số sau:
T=1.2.4+2.3.5+3.4.6+...+100.101.103 / 1.12+ 2.32+ 3.42+...+ 100.302
Rút gọn phân số T=\(\frac{1.2.4+2.3.5+3.4.6+...+100.101.103}{1.2^2+2.3^2+3.4^2+...+100.101^2}\)
\(p=\frac{1.2.4+2.3.5+3.4.6+...+100.101.103}{1.2^2+2.3^2+3.4^2+...+100.101^2}\)Rút gọn P
Tử số = \(1.2.4+2.3.5+3.4.6+...+100.101.103\)
\(=1.2.\left(3+1\right)+2.3.\left(4+1\right)+3.4.\left(5+1\right)+...+100.101.\left(102+1\right)\)
\(=1.2.3+1.2+2.3.4+2.3+3.4.5+3.4+...+100.101.102+100.101\)
\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)+\left(1.2+2.3+3.4+...+100.101\right)\)
Mẫu số = \(1.2^2+2.3^2+3.4^2+...+100.101^2\)
\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+100.101.\left(102-1\right)\)
\(=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+100.101.102-100.101\)
\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)-\left(1.2+2.3+3.4+...+100.101\right)\)
đặt \(A=1.2.3+2.3.4+3.4.5+...+100.101.102\) và \(B=1.2+2.3+3.4+...+100.101\)
bạn tự tính : \(A=\frac{100.101.102.103}{4}=25.101.102.103\); \(B=\frac{100.101.102}{3}=100.101.34\)
rồi thay vào tìm P=\(\frac{A+B}{A-B}\)
Tính B=1.2.4+2.3.5+3.4.6+....+n(n+1)(n+3)
Tính B=1.2.4+2.3.5+3.4.6+....+n(n+1)(n+3)
tinh
A=1.1!+2.2!+3.3!+.........+n.n!
B=1.2.4+2.3.5+3.4.6+4.5.7+.....+n.(n+1).(n+3)
A=1.22+2.32+3.42+...+2017.20182
D=1.22+2.32+3.42+...........+99.1002
Các số có tổng từ 1->100 có tổng là:2600
Có 200 số 2 nên ta lấy
2600.200=520 000
=>D=520 000
Tính : Q = 1.22 + 2.32 + 3.42 + …+ 19. 202
Q=1.2.(3-1)+2.3.(4-1)+3.4.(5-1)+...+19.20.(21-1)=
=(1.2.3+2.3.4+3.4.5+...+19.20.21)-(1.2+2.3+3.4+...+19.20)
Đặt
A=1.2.3+2.3.4+3.4.5+...+19.20.21
4A=1.2.3.4+2.3.4.4+3.4.5.4+...+19.20.21.4=
=1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+19.20.21.(22-18)=
=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-18.19.20.21+19.20.21.22=
=19.20.21.22
\(A=\dfrac{19.20.21.22}{4}=5.19.21.22\)
Đặt
B=1.2+2.3+3.4+...+19.20
3B=1.2.3+2.3.3+3.4.3+...+19.20.3=
=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)=
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-18.19.20+19.20.21=
=19.20.21
\(B=\dfrac{19.20.21}{3}=7.19.20\)
Q=A-B
tính B=1.2.4+2.3.5+...+n(n+1)(n+3)