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Đặt \(A=1.2+2.3+3.4+...+99.100\)
\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)
\(3A=99.100.101\)
\(3A=999900\)
\(A=999900:3\)
\(A=333300\)
3S= 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
Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101 3S = 3.33.100.101
S=33.100.101= 333300
Đặt
S= 1.2 + 2.3 + 3.4 + ...+ 99.100
3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101 3S = 3.33.100.101
S=33.100.101= 333300
s=1.2+2.3+3.4+...+99.100
=>3s=1.2.3+2.3.3+3.4.3+...+99.100.3
=1.2.3+2.3.(4-1)+...+99.100.(101-98)
=1.2.3-1.2.3+2.3.4-2.3.4+...-98.99.100+99.100.101
=99.100.101
=>s=99.100.101/3=333300
\(\text{ta có: }\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
...........................
\(\frac{1}{39.40}=\frac{1}{39}-\frac{1}{40}\)
Đồng nhất 2 vế ta có:
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{39}-\frac{1}{40}=\frac{1}{2}-\frac{1}{40}=\frac{19}{40}\)
Ta có :
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)
\(=\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)
\(=\)\(\frac{1}{2}-\frac{1}{40}\)
\(=\)\(\frac{20}{40}-\frac{1}{40}\)
\(=\)\(\frac{19}{40}\)
Vậy \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}=\frac{19}{40}\)
sorry , ko để ý,
Ta có A =1.2 + 2.3 + 3.4 + ...+ 98.99
B = 1^2 + 2^2 + 3^2 +...+98^2 = 1.1+2.2+3.3+...+98.98
Suy ra: A-B= (1.2 + 2.3 + 3.4 + ...+ 98.99) - (1.1+2.2+3.3+...+98.98)
= (1.2-1.1) + (2.3-2.2) + (3.4-3.3) +...+ (98.99-98.98)
= 1(2-1) + 2(3-2) + 3(4-3) +...+ 98(99-98)
= 1.1 + 2.1 + 3.1 +...+ 98.1
= 1+ 2+ 3+...+ 98 = [98.(98+1)]/2= 98.99/2 = 4851
A = 1.2 + 2.3 + 3.4 + ... + 98.99
A x 3 =1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3
A x 3 = 1.2.3 + 2.3.(4-1 ) + 3.4.(5-2 )+...+98.99.(100-97)
A x 3 = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99
A x 3 = 98.99.100
=> A = 98.99.100:3
=> A = 323400
3B = 1.2.3 + 2.3.3 + 3.3.4 + .... + 3.99.100
Đặt M = 1.2.3 + 2.3.4 + 3.4.5 + .... + 99.100.101
=> M - 3A = 1.2.3 - 1.2.3 + 2.3.(4-3) + 3.4 ( 5-3) + .... + 99.100 ( 101 -3)
= 1.2.3 + 2.3.4 + .... + 98.99.100
=> M -3D = M - 99.100.101
=> D = 99.100.101/3 = 333300
Bạn nhân cả 2 vế với 3 nhé
3D=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3D=99.100.101
D=99.100.101:3=333300
Đặt A = 1.2+2.3+3.4+........+99.100
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + 99.100.(101 - 98)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101
=> 3A = 99.100.101
=> A = 99.100.101/3
=> A = 333300