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Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)
\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)
\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)
\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)
\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)
\(=49\cdot50\cdot67+2500=166650\)
Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)
\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)
\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)
\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)
\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)
\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)
\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)
Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)
\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)
\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)
\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)
\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)
\(=50\cdot101\cdot70=3500\cdot101=353500\)
Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)
\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)
\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)
\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)
\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)
\(=49\cdot33\cdot203=328251\)
A = 1×3+3×5+5×7+...+ 97×99+99×101
6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6
6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)
6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101
6A= 1×3+99×101×103
6A= 1029900
A= 171650
Đặt A = 1.4 + 2.5 + 3.6 + ... + 100.103
= 1.(2 + 2) + 2.(3 + 2) + 3.(4 + 2) +.... + 100.(101 + 2)
= 1.2 + 2.3 + 3.4 + ... + 100.101 + (1.2 + 2.2 + 3.2 + ... + 100.2)
= 1.2 + 2.3 + 3.4 + ... + 100.101 + 2(1 + 2 + 3 + .... + 100)
= 1.2 + 2.3 + 3.4 + .... + 100.101 + 2.100.(100 + 1) : 2
= 1.2 + 2.3 + 3.4 + ... + 100.101 + 10100
Đặt B = 1.2 + 2.3 + 3.4 + .... + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + .... + 100.101.3
=> 3B = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101.(102 - 99)
=> 3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 100.101.102 - 99.100.101
=> 3B = 100.101.102
=> B = 343400
Khi đó A = 343400 - 10100 = 333300
3F= 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)
=>F
H=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
=> 4H=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)((n+3)-(n-1))
=1.2.3.4-0.1.2.3+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)