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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
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Ta thấy: 1.4 = 1.(1 + 3)
2.5 = 2.(2 + 3)
3.6 = 3.(3 + 3)
4.7 = 4.(4 + 3)
…….
n(n + 3) = n(n + 1) + 2n
Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n
C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n
C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)
⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n)
3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)
3C = n(n + 1)(n + 2) +
⇒ C = +
=
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\(1\cdot4+2\cdot5+...+100\cdot103\)
\(=1\left(1+3\right)+2\left(2+3\right)+...+100\left(100+3\right)\)
\(=3\left(1+2+...+100\right)+\left(1^2+2^2+...+100^2\right)\)
\(=\dfrac{3\cdot100\cdot101}{2}+\dfrac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}\)
\(=3\cdot50\cdot101+338350=353500\)