1. Let's restate the problem correctly based on your clarification. We want to analyze the inequality involving the summation on the left side: $$\sum_{t=1}^k \binom{t}{\lfloor \frac{t}{2} \rfloor} \cdot 2$$ and on the right side: $$\binom{n}{2}$$ 2. The problem is to write the expression correctly with the summation from $t=1$ to $k$, where each term is $2$ times the binomial coefficient $\binom{t}{\lfloor \frac{t}{2} \rfloor}$, and the right side is the binomial coefficient $\binom{n}{2}$. 3. The corrected inequality expression is: $$ \sum_{t=1}^k 2 \cdot \binom{t}{\lfloor \frac{t}{2} \rfloor} \leq \binom{n}{2} $$ 4. Here, $\binom{t}{\lfloor \frac{t}{2} \rfloor}$ is the binomial coefficient "t choose floor of t over 2", and $\binom{n}{2}$ is "n choose 2". 5. This expression correctly represents the inequality with the summation and binomial coefficients as you described. Final expression: $$ \sum_{t=1}^k 2 \cdot \binom{t}{\lfloor \frac{t}{2} \rfloor} \leq \binom{n}{2} $$