By Abraham P Hillman
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R&1 r r (1) This formula provides an efficient method of generating successive lines of the Pascal Triangle, but the method is not the best one if we want only the value of a single binomial coefficient for a large n, such as 100 . 3 We therefore seek a more direct approach. It is clear that the binomial coefficients in a diagonal adjacent to a diagonal of 1's are the 40 n 1 numbers 1, 2, 3, ... ; that is, ' n. Now let us consider the ratios of binomial coefficients to the previous ones on the same row.
D) Ln%10 ' 11Ln%5 % Ln. 20. State an analogue of Example 4 for the Fibonacci numbers instead of the Lucas numbers and use it to prove analogues of the formulas of Problem 19. 21. In each of the following parts, evaluate the expression for some small values of n, use this data to make a conjecture, and then prove the conjecture true for all integers n. 2 (a) Fn%1 & Fn Fn%2. 2 (b) 2 Fn%2 & Fn%1 Fn . (c) Fn-1 + Fn+1. 22. Discover and prove formulas similar to the first two parts of the previous problem for the Lucas numbers.
K & 1. for r ' 0, 1, 2, ... 47 &4 r 10. Prove that r%3 3 ' (&1)r for r ' 0, 1, 2, ... 11. Let m be a positive integer and r a non-negative integer. Express n k binomial coefficient &m r in terms of a with 0 # k # n. 12. In the original definition of n r as a binomial coefficient, it was clear that it was always an integer. Explain why this is still true in the extended definition. n a 13. Show that n&a n! (n & a & b)! a $ 0, b $ 0, and n $ a % b. 14. Given that n = a + b + c + d and that a, b, c, and d are non-negative integers, show that n a n&a b n&a&b c n&a&b&c d 15.