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\(a,\)Biết \(B=\frac{100.101}{2}=50.101\)
\(A=1^3+2^3+3^3+...+99^3+100^3\)
Xét \(A=\left(1^3+100^3\right)+\left(2^3+99^3\right)+...+\left(49^3+52^3\right)+\left(50^3+51^3\right)\)
\(\Rightarrow A=101.\left(1+100+100^2\right)+101.\left(2^2+2.99+99^2\right)+...+101\left(50^2+50.51+51^2\right)\)
\(\Rightarrow A=101\left(1+100+100^2+2^2+2.99+99^2+...+50^2+50.51+51^2\right)⋮101\)
Xét\(A=\left(1^3+99^3\right)+\left(2^3+98^3\right)+...+\left(49^3+51^3\right)+50^3\)
\(\Rightarrow A=100\left(1^2+1.99+99^2\right)+100\left(2^2+2.98+98^2\right)+...+100\left(49^2+49.51+51^2\right)+100.50.25⋮50\)
Vậy \(A⋮101.50=5050=B\)
Làm tương tự với câu b
Ta có: \(\dfrac{n^3-1}{n^3+1}=\dfrac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\dfrac{\left(n-1\right)[\left(n+0,5\right)^2+0,75]}{\left(n+1\right)[\left(n-0,5\right)^2+0,75]}\)
Thay vào M ta có:
\(M=\dfrac{2,5^2+0.75}{3.\left(1,5^2+0,75\right)}.\dfrac{2.\left(3,5^2+0,75\right)}{4.\left(2,5^2+0,75\right)}...\dfrac{99[\left(100,5\right)^2+0,75]}{101.[\left(99,5\right)^2+0,75}\)
\(=\dfrac{1.2.3...99}{3.4.5...101}.\dfrac{\left(2,5^2+0,75\right).\left(3,5^2+0,75\right)...[\left(100,5\right)^2+0,75]}{\left(1,5^2+0,75\right).\left(2,5^2+0,75\right)...[\left(99,5\right)^2+0,75]}\)\(=\dfrac{1.2}{100.\left(101\right)}.\dfrac{\left(100,5\right)^2+0,75}{1,5^2+0,75}=\dfrac{2}{3}.\dfrac{\left(100^2+100+1\right)}{3.100.101}>\dfrac{2}{3}\left(đpcm\right)\)
Ta có : \(B=\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)=101.50\)
Ta lại có : \(A=\left(1^3+100^3\right)+\left(2^3+99^3\right)+...+\left(50^3+51^3\right)\)
\(=\left(1+100\right)\left(1^2+100+100^2\right)+\left(2+99\right)\left(2^2+2.99+99^2\right)+...+\left(50+51\right)\left(50^2+50.51+51^2\right)\)
\(=101.\left(1^2+100+100^2+2^2+2.99+99^2+...+50^2+50.51+51^2\right)\) chia hết cho 101 (1)
Lại có : \(A=\left(1^3+99^3\right)+\left(2^3+98^3\right)+...+\left(50^3+100^3\right)\)
Mỗi số hạng trong ngoặc đều chia hết cho 50 nên A chia hết cho 50 (2)
Từ (1) và (2) suy ra A chia hết cho 101 và 50 nên A chia hết cho B (đpcm)
A = -12 + 22 - 32 + 42 - ... - 992 + 1002
A = 1002 - 992 + ... + 42 - 32 + 22 - 12
A = (100 + 99).(100 - 99) + ... + (4 + 3).(4 - 3) + (2 + 1).(2 - 1)
A = 100 + 99 + ... + 4 + 3 + 2 + 1
\(A=\frac{\left(1+100\right).100}{2}=101.50=5050\)
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
2B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)...(332 + 1)
2B = (32 - 1)(32 + 1)(34 + 1)...(332 + 1)
2B = (34 - 1)(34 + 1)...(332 + 1)
2B = 364 - 1
\(B=\frac{3^{64}-1}{2}\)
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 \(B=E+F\)
\(\Rightarrow B=5050+25497450\)
\(\Rightarrow B=25502500\)