Question

Suppose that 1 ≤ k < n. Prove that the part k occurs a total of
(n − k + 3)2n−k−2
times among all the 2n−1
compositions of n. For
example, if n = 4 and k = 2, then the part 2 occurs once in 2 + 1 + 1,
1 + 2 + 1, and 1 + 1 + 2, and twice in 2 + 2, for a total of 5 =
(4 − 2 + 3)24−2−2
times.

Answer 1

id say use induction

Answer 2

(4-2+3)*2*4-2-2
(2+3)*8
5*8
40 not 5

Answer 3

well actually (4-2+3)*2*4-2-2
(2+3)*8-4
5*8-4
36 not 5

Answer 4

something is wrong with the (n-k+3)2n-k-2