Consider the following three claims — I.(n+k)m = Θ(nm) where k and m are…

Consider the three claims:

I. (n + k)^m = Θ(n^m), where k and m are constants

II. 2^(n+1) = O(2^n)

III. 2^(2n) = O(2^n)

Evaluation:

Statement I: TRUE

Since k and m are constants, (n + k)^m can be expanded using the binomial theorem:

(n + k)^m = n^m + m·k·n^(m-1) + lower order terms.

The highest power term is n^m, and lower-order terms do not affect asymptotic growth. Hence, (n + k)^m grows at the same rate as n^m, so it is Θ(n^m).

Statement II: TRUE

We have 2^(n+1) = 2 · 2^n. This is just a constant multiple of 2^n. In Big-O notation, constant factors are ignored. Therefore, 2^(n+1) = O(2^n).

Statement III: FALSE

We have 2^(2n) = (2^2)^n = 4^n. This grows much faster than 2^n. {i.e. a^(bc) = (a^b)^c}

Consider the ratio: 2^(2n) / 2^n = 2^n → ∞ as n → ∞.

Since it is not bounded above by a constant multiple of 2^n, the Big-O relation does not hold.

Final Conclusion:

Statement I and II are correct, while Statement III is incorrect.