By Abraham P Hillman

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N % m) . m % 1 Prove it for general m. * 33. Prove that n5 - n is an integral multiple of 30 for all integers n. * 34. Prove that n7 - n is an integral multiple of 42 for all integers n. * 35. Show that every integer from 1 to 2n+1 - 1 is expressible uniquely as a sum of distinct powers of 2 chosen from 1, 2, 22, ... , 2n. 36. Show that every integer s from & * 3n%1 & 1 3n%1 & 1 to has a unique expression of the 2 2 form s ' c0 % 3c1 % 32c2 % ... , cn is 0, 1, or -1. 39 Chapter 6 THE BINOMIAL THEOREM n r In Chapter 1 we defined as the coefficient of an-rbr in the expansion of (a + b)n, and tabulated these coefficients in the arrangement of the Pascal Triangle: n Coefficients of (a + b)n 0 1 1 1 2 1 3 1 4 1 5 1 6 ...

They can be arranged in the following ways: abcd bacd cabd dabc abdc badc cadb dacb acbd bcad cbad dbac acdb bcda cbda dbca adbc bdac cdab dcab adcb bdca cdba dcba 53 Rather than write them all out, if we are only interested in the number of arrangements, we may think of the problem thus: We have four spaces to fill. If we put, for example, the b in the first, we have only the a, c, and d to choose from in filling the remaining three. And if we put the d in the second, we have only a and c for the remaining; and so forth.

C4 = 3a + 2. c5 = 5a + 3. c6 = 8a + 5. 42. For the sequence c0, c1, ... of the previous problem, express c12 in the form aFu + Fv, where Fu and Fv are Fibonacci numbers, and conjecture a similar expression for cm. 27 * 43. In the sequence 1/5, 3/5, 4/5, 9/10, 19/20, 39/40, ... each succeeding term is the average of the previous term and 1. Thus: 3 1 1 4 1 3 9 1 4 ' % 1 , ' % 1 , ' % 1 , ÿ. 5 2 5 5 2 5 10 2 5 (a) Show that the twenty-first term is 1 & 1 . 5@218 (b) Express the nth term similarly.

### Algebra through problem solving by Abraham P Hillman

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