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The number of nodes per level grows as a power of 2 because each recursive call splits into exactly two subcalls. Starting with 1 node at the root (level 1, $2^0$), the second level has 2 nodes ($2^1$), the third has 4 nodes ($2^2$), and the fourth has 8 nodes ($2^3$).

Conditions: The recurrence relation is $f(n) = f(n-1) + f(n-2)$, producing a binary tree.; Levels are numbered starting from 1.; The tree is counted as a full binary tree for theoretical estimation.

The number of nodes per level grows as a power of 2 because each recursive call splits into exactly two subcalls. Starting with 1 node at the root (level 1, $2^0$), the second level has 2 nodes ($2^1$), the third has 4 nodes ($2^2$), and the fourth has 8 nodes ($2^3$).

Conditions: The recurrence relation is $f(n) = f(n-1) + f(n-2)$, producing a binary tree.; Levels are numbered starting from 1.; The tree is counted as a full binary tree for theoretical estimation.

The video mentions this specific finite geometric series sum to provide a concrete example that illustrates the growth trend of summing the nodes across all levels of the recursion tree. It demonstrates that the total sum is dominated by the highest power, supporting the intuitive derivation of the $O(2^n)$ time complexity.

Conditions: The sum starts from $2^0$.; The common ratio is 2.; The last term is $2^6$.

The video mentions this specific finite geometric series sum to provide a concrete example that illustrates the growth trend of summing the nodes across all levels of the recursion tree. It demonstrates that the total sum is dominated by the highest power, supporting the intuitive derivation of the $O(2^n)$ time complexity.

Conditions: The sum starts from $2^0$.; The common ratio is 2.; The last term is $2^6$.