## Task A-4.6.

Calculate

$$
\sum_{k=1}^{1011}\binom{2022}{2 k-1} 2^{2 k-1}
$$
## Second Solution.

We solve the problem using a combinatorial interpretation.

The binomial coefficient $\binom{2022}{2 k-1}$ is the number of ways to choose a subset with $2 k-1$ elements from a set of 2022 elements.

The number $2^{2 k-1}$ can be interpreted as the number of subsets of a set with $2 k-1$ elements.

Therefore, the given sum is equal to the number of ways to choose: a number $k$ between 1 and 1011, a subset $B$ of the set $S=\{1,2, \ldots, 2022\}$ with $2 k-1$ elements, and a subset $A$ of the set $B$.

Thus, we are choosing pairs $(A, B)$ such that $A \subseteq B \subseteq S$ and $B$ has an odd number of elements.

![](https://cdn.mathpix.com/cropped/2024_05_30_21f39afeb1dfc245d841g-26.jpg?height=314&width=417&top_left_y=734&top_left_x=748)

The number of pairs $(A, B)$ such that $A \subseteq B \subseteq S$ (without the condition that $B$ has an odd number of elements) is equal to $3^{2022}$ because each such pair is uniquely determined by the placement of each element of the set $S$ into one of the disjoint sets $A, B \backslash A$, and $S \backslash B$.

We will show that all pairs except $(A, B)=(S, S)$ can be paired such that pairs $\left(A, B^{\prime}\right)$ and $\left(A, B^{\prime \prime}\right)$ are paired if and only if there is exactly one element that appears in one of the sets $B^{\prime}$ and $B^{\prime \prime}$, but not in the other. In this way, one pair with an odd number of elements in $B$ will be paired with one pair in which $B$ has an even number of elements. Therefore, the number of desired pairs is $\frac{1}{2} \cdot\left(3^{2022}-1\right)$.

Indeed, for each pair $(A, B)$, let $k$ be the largest element of the set $S$ that is not in $A$. If $k \in B$, then we pair $(A, B)$ with $(A, B \backslash\{k\})$, and if $k \notin B$, then we pair $(A, B)$ with $(A, B \cup\{k\})$. In other words, we move the largest element $k$ that is in either $B \backslash A$ or $S \backslash B$ between these sets. In this way, we can pair all pairs except $(A, B)=(S, S)$, because only in this case are there no elements in $S$ that are not in $A$.

![](https://cdn.mathpix.com/cropped/2024_05_30_21f39afeb1dfc245d841g-26.jpg?height=311&width=417&top_left_y=1849&top_left_x=748)