Nhưng chúng ta có thể dễ dàng mâu thuẫn với bất bình đẳng n! nn / 2 mà chúng ta bắt nguồn trong (4,22). Là những số nguyên tố vô cùng nhiều, vẫn còn. Chúng tôi thậm chí có thể tăng cường lập luận này để có được một thô ràng buộc về n (n), | BASIC PRACTICE 179 have the form Rk in the resulting expression somewhat as we did in the perturbation method of Chapter 2 Anyway those of us who ve done warmup exercise 4 know it. In the next-to-last step we ve used the formula m l m which we know is true when m 0. This derivation is valid for m 2. From this recurrence we can generate values of Rln quickly and we soon perceive that the sequence is periodic. Indeed Rm I I 0 -1 -1 0 0 I 2 if m mod 6 7 3 4 The proof by induction is by inspection. Or if we must give a more academic proof we can unfold the recurrence one step to obtain Rm - Rm-2 Rm-3 - Rm-2 - Rm-3 whenever m 3. Hence Rm Rm-6 whenever m 6. Finally since Qn R2 we can determine Qn by determining 2 mod 6 and using the closed form for Rm. When n 0 we have 2 mod 6 1 after that we keep multiplying by 2 mod 6 so the pattern 2 4 repeats. Thus i R1 1 if n. 0 Qn Ri R2 0 if n is odd I R4 -1 if n 0 is even. This closed form for Qn agrees with the first four values we calculated when we started on the problem. We conclude that Q10Ũ0000 R4 -1. 180 BINOMIAL COEFFICIENTS Problem 4 A sum involving two binomial coefficients. Our next task is to find a closed form for Ei trt k 1 n k I integers m n z 0. m -n-1 k 0 v Wait a minute. Where s the second binomial coefficient promised in the title of this problem And why should we try to simplify a sum we ve already simplified This is the sum s from Problem 2. Well this is a sum that s easier to simplify if we view the summand as a product of two binomial coefficients and then use one of the general identities found in Table 169. The second binomial coefficient materializes when we rewrite k as And identity is the one to apply since its index of summation appears in both upper indices and with opposite signs. But our sum isn t quite in the correct form yet. The upper limit of summation should be m 1 if we re to have a perfect match with . No problem the terms for n k m 1 arc zero. So we can plug in with 1 m n q m -
