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Đặt \(A=1.4+2.5+3.6+...+100.103\)
\(=1\left(2.2\right)+2\left(3+2\right)+3\left(4+2\right)+...+100\left(101+2\right)\)
\(=1.2+2.3+3.4+...+100.101+\left(1.2+2.2+3.2+...+100.2\right)\)
\(=1.2+2.3+3.4+...+100.101+2\left(1+2+3+...+100\right)\)
\(=1.2+2.3+3.4+...+100.101+2.100\left(100+1\right):2\)
\(=1.2+2.3+3.4+...+100.101+10100\)
Đặt \(B=1.2+2.3+3.4+...+100.101\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+100.101.3\)
\(\Rightarrow3B=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)\)
\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)
\(\Rightarrow3B=100.101.102\)
\(\Rightarrow B=343400\)
Khi đó \(A=343400=10100=333300\)
Đặt A = 1.4 + 2.5 + 3.6 + 4.7 + ... + 100.103
3A = 3.(1.2 + 2.3 + 3.4 + ... + 100.101] + 3.(2 + 4 + 6 + ... + 200)
= 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3 + 3.(2 + 4 + 6 + ... + 200)
\(\Rightarrow\) A = 100.101.105:3 = 353500
1.4+2.5+3.6+...+99.102=1(2+2)+2(3+2)+3(4+2)+...99(100+2)
=1.2+1.2+2.3+2.2+...+99.100+99.2
=(1.2+2.3+...+99.100)+2(1+2+...+99)
A=1.2+2.3+3.4+...+99.100(cho A la ten bieu thuc nay)
3A=1.2(3-0)+2.3(4-1)+...+99.100(101-98)
=(1.2.3+2.3.4+...+99.100.101)-(1.2.3+2.3.4+3.4.5+...+98.99.100
=99.100.101=>A=\(\frac{99.100.101}{3}\)=33330
2.(1+2+...99)
=2(100.99:2)=2.4950=9900
33330+9900=343200
vay...
`S = 1.4+2.5 + 3.6 +...+ 100.103`
`S = 1 . (2+2) + 2 (3+2) + 3 . (4+2) + ... + 100 . (101 + 2) `
`S = 1.2 + 2 . 1 + 2.3 + 2.2 + 3.4 + 2.3 + ... + 100 . 101 + 2. 100`
`S = 1.2 + 2.3+ 3.4 +... + 100.101 + 2.1 + 2.2 + 2.3 + ... + 2.100`
`S = 1.2 + 2.3+ 3.4 +... + 100.101 + 2(1+2+3+...+100) `
`S = 1.2 + 2.3+ 3.4 +... + 100.101 + 2 (100+1) . [(100-1):1+1] : 2`
`S = 1.2 + 2.3+ 3.4 +... + 100.101 + 10100`
Đặt `A = 1.2 + 2.3+ 3.4 +... + 100.101`
`3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + .... + 100 . 101. (102 - 99)`
`3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102 -99.100.101`
`3A = 100.101.102`
`A = 343400`
`S = A + 10100`
`= 343400 + 10100`
`= 353500`