I have recently begun my studies in an upper-level discrete mathematics course. So far, I have quite enjoyed all of the lectures.

I recently came across an exercise on a supplemental homework requesting me to compute a trio of three binomial sums containing floor functions:

$$\sum_{k=0}^{\left\lfloor\frac{n}{3}\right\rfloor} {n \choose 3k}$$ $$\sum_{k=0}^{\left\lfloor\frac{n}{3}\right\rfloor} {n \choose 3k+1}$$ $$\sum_{k=0}^{\left\lfloor\frac{n}{3}\right\rfloor} {n \choose 3k+2}$$

About an hour ago, I had not even heard of ceiling and floor functions.

I have tried to work the first sum out, but I really am not sure how to continue with these floor functions, nor am I sure how to truly approach these sums.

Any and all help would be greatly appreciated! Thank you all for taking the time to read my post.

  • $\begingroup$ The floors are not actually a problem here. What do you know about binomials ? do you know z-Transform ? $\endgroup$ – G Cab Jan 22 at 23:52

Hint: Let $a_n$ be your first summation, and $b_n,c_n$ be the second and third. Start by computing these exactly for small values of $n$. You should start to notice a general pattern emerging. Describe the pattern exactly, then prove that it holds in general using induction on $n$. Use the base cases $$ a_0 = 1, b_0=0,c_0=0 $$ and the rules $$ a_{n+1}=a_n+c_n,\qquad b_{n+1}=b_n+a_n,\qquad c_{n+1}=c_n+b_n $$ The rule $a_{n+1}=a_n+c_n$ can be proven by applying Pascal's identity to each summand in $a_{n+1}$, and then splitting into two summations, which will be exactly $a_n$ and $c_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.