Let f(n) = 2^{n+1} and g(n) = 1 + (n+1)2^n fo | vedclass

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(B) Consider the difference $g(n) - f(n) = 1 + (n+1)2^n - 2^{n+1}$.Since $2^{n+1} = 2 \cdot 2^n$, we have:$g(n) - f(n) = 1 + (n+1)2^n - 2 \cdot 2^n$$g

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$f(n) < g(n)$

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(B) Consider the difference $g(n) - f(n) = 1 + (n+1)2^n - 2^{n+1}$.Since $2^{n+1} = 2 \cdot 2^n$, we have:$g(n) - f(n) = 1 + (n+1)2^n - 2 \cdot 2^n$$g(n) - f(n) = 1 + (n+1-2)2^n$$g(n) - f(n) = 1 + (n-1)2^n$.For all $n \in N$, $n \geq 1$, which implies $(n-1) \geq 0$.Since $2^n > 0$, the term $(n-1)2^n \geq 0$.Therefore, $1 + (n-1)2^n > 0$, which means $g(n) - f(n) > 0$ or $g(n) > f(n)$.

Let $f(n) = 2^{n+1}$ and $g(n) = 1 + (n+1)2^n$ for all $n \in N$. Then:

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