Try It Now

1. By the recursive definition of binomial coefficients, \(\binom{7}{2} = \binom{6}{2} + \binom{6}{1}\). Continue expanding \(\binom{7}{2}\) to express it in terms of quantities defined by the basis. Check your result by applying the factorial definition of \(\binom{n}{k}\).
   
   
3. Let p(x) = x⁵ + 3x⁴− 15x³ + x − 10.
   1. Write p(x) in telescoping form.
   2. Use a calculator to compute p(3) using the original form of p(x).
   3. Use a calculator to compute p(3) using the telescoping form of p(x).
   4. Compare your speed in parts b and c.
      
      
5. What is wrong with the following definition of f : ℝ → ℝ? f (0) = 1 and f(x) = f(x/2)/2 if x ≠ 0.

Source: Al Doerr and Ken Levasseur, http://faculty.uml.edu/klevasseur/ads-latex/ads.pdf
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. Answer:
   
   \(\begin{align*} \binom{7}{2} &= \binom{6}{2} + \binom{6}{1}\\   &= \binom{5}{2} + \binom{5}{1} + \binom{5}{1} + \binom{5}{0} \\   &= \binom{5}{2} + 2\binom{5}{1} + 1\\   &= \binom{4}{2} + \binom{4}{1} + 2(\binom{4}{1} + \binom{4}{0}) + 1\\   &= \binom{4}{2} + 3(\binom{4}{1}) + 3 \\   &= \binom{3}{2} + \binom{3}{1} + 3(\binom{3}{1} + \binom{3}{0}) + 3\\   &= \binom{3}{2} + 4\binom{3}{1} + 6 \\   &= \binom{2}{2} + \binom{2}{1} + 4(\binom{2}{1} + \binom{2}{0}) + 6 \\   &= 5\binom{2}{1} + 11\\   &= 5(\binom{1}{1} + \binom{1}{0}) + 11\\   &= 21 \end{align*}\)
   
   
3. Answer:
   
   1. p(x) in telescoping form: (((( x + 3)x − 15)x + 0)x + 1)x − 10
   2. p(3) = ((((3 + 3)3 − 15)3 − 0)3 + 1)3 − 10 = 74
      
      
5. Answer: The basis is not reached in a finite number of steps if you try to compute f (x) for a nonzero value of x.