Let $C$ be the radix $3$ $2$-error correcting linear code with $m=4$ check bits. What is the greatest possible number of information bits $k$ in $C$?
I have been instructed to use the Sphere-Packing Bound: $$|C|\sum_{i=0}^t{n\choose i}(r-1)^i\leq r^n. $$ As $|C$| is linear, this implies $|C|=r^k$ ($r$ denotes the radix). It is known that $m=n-k$. Thus, \begin{align} 3^k\sum_{i=0}^2{k+4\choose i}2^i\leq 3^{k+4} \\ 3^k\left({k+4\choose 0}+2{k+4\choose 1}+4{k+4\choose 2}\right)\leq 3^{k+4} \\ 1+2(k+4)+4{k+4\choose 2}\leq 3^{4}. \\ (k+4)+2{k+4\choose 2}\leq 40 \end{align} My question is, how do you simplify this last line to find an upper bound of $k$? I am very rusty.
You have $$(k+4)+2((k+4)(k+3)/2)=(k+4)^2\leq 40,$$ by definition of the binomial ceefficient, which gives $k=2,$ as the largest integer satisfying the inequality.