It's a famous result that a symmetric random walk in 1-d comes back to the origin an infinite number of times, so does a symmetric random walk in 2-d. But a symmetric random walk in 3-d comes back to the origin only a finite number of times. For the 1-d and 2-d cases, this is show in the book Introduction to Probability models by Sheldon Ross, example 4.18.
We know that for the 3-d case, the probability of coming back to the origin is 37% (https://mathworld.wolfram.com/PolyasRandomWalkConstants.html).
This begs the question, what is the expected number of times a 3-d random walk will come back to the origin?
My approach: take the summation in the question: Summation coming about in the process of solving a 3d random walk $\sum\limits_{i=0}^n {n \choose i}^2 {2(n-i)\choose (n-i)}$ and numerically calculate it for a large number of terms. I'm wondering if there is a better way or a purely mathematical solution.