Answer:
If (K + 1)^2 = 2n + (n^2)/2, then the answer is Yes, else No.
Step-by-step explanation:
Sum of first K + 1 odd numbers (S): 1 + 3 + 5 + ... + (2K - 3) + (2K - 1) + (2K + 1).
(1 + 3 + 5 + ... + (2K - 3) + (2K - 1) + (2K + 1)) + ((2K + 1) + (2K - 1) + (2K - 3) + ... + 5 + 3 + 1 ) = (2K + 2) + (2K + 2) + (2K + 2) + ... + (2K + 2) + (2K + 2) + (2K + 2) (repeat K + 1 times) = (2K + 2)(K + 1) = 2S
2S = (2K + 2)(K + 1)
S = (2K + 2)(K + 1)/2
S = (2K^2 + 4K + 2)/2
S = K^2 + 2K + 1
S = (K + 1)^2
(4n + n^2)/2 = 2n + (n^2)/2
