a) \(A=1^2+2^2+3^2+...+100^2\)
\(A=1.1+2.2+3.3+...+100.100\)
\(A=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+100.\left(101-1\right)\)
\(A=1.2-1+2.3-2+3.4-3+...+100.101-100\)
\(A=\left(1.2+2.3+3.4+...+100.101\right)-\left(1+2+3+...+100\right)\)
Đặt \(C=1.2+2.3+3.4+...+100.101\)và \(D=1+2+3...+100\)
Ta có:
\(C=1.2+2.3+3.4+...+100.101\)
\(3C=1.2.3+2.3.3+3.4.3+...+100.101.3\)
\(3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101\left(102-99\right)\)
\(3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)
\(3C=100.101.102\)
\(3C=1030200\)
\(C=343400\)
Ta lại có:
\(D=1+2+3+...+100\)
\(D=\dfrac{100-1+1}{2}.\left(1+100\right)\)
\(D=50.101\)
\(D=5050\)
\(\Rightarrow A=C-D\)
\(\Rightarrow A=343400-5050=338350\)
b) Ta thấy:
\(\left(n-1\right)n\left(n+1\right)\)
\(=\left(n^2-n\right)\left(n+1\right)\)
\(=\left(n^2-n\right)n+n^2-n\)
\(=n^3-n^2+n^2-n\)
\(=n^3-n\)
\(\Rightarrow n^3=\left(n-1\right)n\left(n+1\right)+n\left(1\right)\)
Áp dụng (1) vào B ta có:
\(B=1^3+2^3+3^3+...+100^3\)
\(B=\left(1-1\right)1\left(1+1\right)+1+\left(2-1\right)2\left(2+1\right)+2+...+\left(100-1\right)100\left(100+1\right)\)
\(B=1+2+1.2.3+3+2.3.4+...+100+99.100.101\)
\(B=\left(1+2+3+...+100\right)+\left(1.2.3+2.3.4+...+99.100.101\right)\)
Đặt \(E=1+2+3+...+100\)và \(F=1.2.3+2.3.4+...+99.100.101\)
Ta có:
\(E=1+2+3+...+100\)
\(E=\dfrac{100-1+1}{2}.\left(1+100\right)\)
\(E=50.101=5050\)
Ta lại có:
\(F=1.2.3+2.3.4+...+99.100.101\)
\(4F=1.2.3.4+2.3.4.4+...+99.100.101.4\)
\(4F=1.2.3.4+2.3.4\left(5-1\right)+...+99.100.101\left(102-98\right)\)
\(4F=1.2.3.4+2.3.4.5-1.2.3.4+...+99.100.101.102-98.99.100.101\)
\(4F=99.100.101.102\)
\(4F=101989800\)
\(F=25497450\)
Vì \(B=E+F\)
\(\Rightarrow B=5050+25497450\)
\(\Rightarrow B=25502500\)
