Solutions

  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*}

  2. 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

  3. 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.