For every real number x, define [x] to be equal to the greatest integer less than or equal to x. (We call this the "floor" of x.) For example, [4.2] = 4, [5.7] = 5, [-3.4]=-4, [0.4... For every real number x, define [x] to be equal to the greatest integer less than or equal to x. (We call this the "floor" of x.) For example, [4.2] = 4, [5.7] = 5, [-3.4]=-4, [0.4] = 0, and [2] = 2. (a) Determine the integer equal to $\lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + ... + \lfloor \frac{59}{3} \rfloor + \lfloor \frac{60}{3} \rfloor$. (The sum has 60 terms.) (b) Determine a polynomial p(x) so that for every positive integer m > 4, $\lfloor p(m) \rfloor = \lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + ... + \lfloor \frac{m-2}{3} \rfloor + \lfloor \frac{m-1}{3} \rfloor$ (The sum has m - 1 terms.) A polynomial f(x) is an algebraic expression of the form f(x) = $a_nx^n$ + $a_{n-1}x^{n-1}$ + ... + $a_1x + a_0$ for some integer n $\geq$ 0 and for some real numbers $a_n, a_{n-1}, ..., a_1, a_0$. Read more

Steps to Solve

1. Evaluate the floor terms in part (a)

We evaluate the floor of each term in the sum $\lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + ... + \lfloor \frac{59}{3} \rfloor + \lfloor \frac{60}{3} \rfloor$.

2. Group the terms by their floor values

$\lfloor \frac{1}{3} \rfloor = 0, \lfloor \frac{2}{3} \rfloor = 0$. Then $\lfloor \frac{3}{3} \rfloor = 1, \lfloor \frac{4}{3} \rfloor = 1, \lfloor \frac{5}{3} \rfloor = 1$. $\lfloor \frac{6}{3} \rfloor = 2, \lfloor \frac{7}{3} \rfloor = 2, \lfloor \frac{8}{3} \rfloor = 2$. The pattern is that for every increase of 3 in the numerator, the floor increases by 1. We can rewrite our sum, grouping terms, as follows:

$ (\lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor) + (\lfloor \frac{3}{3} \rfloor + \lfloor \frac{4}{3} \rfloor + \lfloor \frac{5}{3} \rfloor) + (\lfloor \frac{6}{3} \rfloor + \lfloor \frac{7}{3} \rfloor + \lfloor \frac{8}{3} \rfloor) + ... + (\lfloor \frac{57}{3} \rfloor + \lfloor \frac{58}{3} \rfloor + \lfloor \frac{59}{3} \rfloor) + \lfloor \frac{60}{3} \rfloor$ which simplifies to $ (0 + 0) + (1 + 1 + 1) + (2 + 2 + 2) + ... + (19 + 19 + 19) + 20 $

3. Calculate the sum in part (a)

We have $2(0) + 3(1) + 3(2) + ... + 3(19) + 20 = 3(1 + 2 + ... + 19) + 20$. The sum of the first n integers is $\frac{n(n+1)}{2}$. Therefore, the sum of integers from 1 to 19 is $\frac{19(20)}{2} = 190$.

Therefore the total sum is $3(190) + 20 = 570 + 20 = 590$.

4. Evaluate the sum in part (b) for different values of m

We want to find a polymial $p(x)$ such that $\lfloor p(m) \rfloor = \lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + ... + \lfloor \frac{m-2}{3} \rfloor + \lfloor \frac{m-1}{3} \rfloor$. Let $S(m) = \lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + ... + \lfloor \frac{m-2}{3} \rfloor + \lfloor \frac{m-1}{3} \rfloor$.

If $m = 5$, then $S(5) = \lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + \lfloor \frac{4}{3} \rfloor = 0 + 0 + 1 + 1 = 2$. If $m = 6$, then $S(6) = \lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + \lfloor \frac{4}{3} \rfloor + \lfloor \frac{5}{3} \rfloor = 0 + 0 + 1 + 1 + 1 = 3$. If $m = 7$, then $S(7) = \lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{3}{3} \rfloor + \lfloor \frac{4}{3} \rfloor + \lfloor \frac{5}{3} \rfloor + \lfloor \frac{6}{3} \rfloor = 0 + 0 + 1 + 1 + 1 + 2 = 5$. If $m = 8$, then $S(8) = \lfloor \frac{1}{3} \rfloor + ... + \lfloor \frac{7}{3} \rfloor = 0 + 0 + 1 + 1 + 1 + 2 + 2 = 7$. If $m = 9$, then $S(9) = \lfloor \frac{1}{3} \rfloor + ... + \lfloor \frac{8}{3} \rfloor = 0 + 0 + 1 + 1 + 1 + 2 + 2 + 2 = 9$. If $m = 10$, then $S(10) = \lfloor \frac{1}{3} \rfloor + ... + \lfloor \frac{9}{3} \rfloor = 0 + 0 + 1 + 1 + 1 + 2 + 2 + 2 + 3 = 12$.

5. Find a pattern for S(m)

The sums $S(5), S(6), S(7), S(8), S(9), S(10)$ are $2, 3, 5, 7, 9, 12$. These don't seem to have an immediately-obvious polynomial pattern

However, we can think about dividing the $m-1$ terms into groups of 3. There are $\lfloor \frac{m-1}{3} \rfloor$ full groups of 3 terms, which contribute $0 + 1 + 2 + ... + (\lfloor \frac{m-1}{3} \rfloor - 1)$ to the sum. But, each such group of 3 actually contribues $0 + 1 + 2 = 3n$, so $0 + 1 + 2$ is the sum of 3 terms where $i/3$ goes from $n$ to $n + 1 -1/3$. Consider $ \sum_{i=1}^{m-1} \lfloor \frac{i}{3} \rfloor$ We can use the following formula: $\sum_{i=1}^{n} \lfloor \frac{i}{k} \rfloor = \frac{(n \mod k)(n - (n \mod k))}{2k} + \frac{k(N)(N-1)}{2k}$ $N = \lfloor \frac{n}{k} \rfloor$. In our problem, $n = m - 1$ and $k = 3$. $N = \lfloor \frac{m-1}{3} \rfloor$. $\sum_{i=1}^{m-1} \lfloor \frac{i}{3} \rfloor = \frac{(m-1 \mod 3)(m-1 - (m-1 \mod 3))}{2 \times 3} + \frac{3 \lfloor \frac{m-1}{3} \rfloor(\lfloor \frac{m-1}{3} \rfloor - 1)}{2 \times 3} $ $ = \frac{(m-1 \mod 3)(m-1 - (m-1 \mod 3))}{6} + \frac{\lfloor \frac{m-1}{3} \rfloor(\lfloor \frac{m-1}{3} \rfloor - 1)}{2} $ When simplified, yields $\frac{(m-1)(m-4)}{6}$. Since $m>4$, the floor is not needed.

$p(x) = \frac{(x-1)(x-4)}{6} = \frac{x^2 - 5x + 4}{6} = \frac{1}{6}x^2 - \frac{5}{6}x + \frac{2}{3}$.