(1) 証明:对任何正整数n n~3+(3/2)n~2+(1/2)n-1都是整数,並且用3除时余2, 証明1.n~3+(2/3)n~2-1=(n(n+1)/2)(2n+1)-1==(2n(2n+1)(2n+2)/8)-3+2。对任何整数n說來,(n(n+1))/2是整数,所以原式为整数。在相鄰的三个整数2n,2n+1,2n+2中至少有一个是3的倍数,因为3与8互質,8除得尽分子,分子提出3后还能被8除尽,所以(2n(2n+1)(2n+2)/8)-3是3的倍数,原式用3除时余2。 証明2.用f(n)代表原式,現在用数学归納法証明我們的問題。在n=1时, (1) Proof: For any positive integer nn~3+(3/2)n~2+(1/2)n-1 is an integer, and divide by 3 to prove that 1.n~3+ ( 2/3) n~2-1=(n(n+1)/2)(2n+1)-1==(2n(2n+1)(2n+2)/8)-3+2. For any integer n, (n(n+1))/2 is an integer, so the primitive is an integer. At least one of the adjacent three integers 2n, 2n+1, 2n+2 is a multiple of 3, because 3 and 8 are relatively prime, 8 divides the numerator, and after the numerator proposes 3, it can also be divided by 8, so (2n(2n+1)(2n+2)/8)-3 is a multiple of 3, and the original formula is divided by 3 and is more than 2. Proof 2. Use f(n) to represent the original, and now use mathematical induction to prove our problem. When n=1,