Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$, where did I go wrong in my induction step?

Question

Can someone help me out with the induction step?

Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$

Base case n=1: $$4^1 = 2^{1*(1+1)} = 2^2$$

Induction step (to show: $2^{(n+1)*(n+2)} = 2^{n^2+3n+2}$ ) :

$$ \prod_{k=1}^{n+1} 4^k = 4^{n+1} * 2^{n*(n+1)} = \frac{1}{2} * 4^{n+1} * 4^{n*(n+1)} = \frac{1}{2} 4^{n^2+n+n+1}= \frac{1}{2} 4^{n^2+2n+1} = 2^{n^2+2n+1}$$

but now we have

$$ 2^{n^2+3n+2} \neq 2^{n^2+2n+1} $$

Where did I go wrong?

Answer 1

The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$ and so

$\prod_{k=1}^n4^k=4^{\sum_{k=1}^nk}=4^{\frac{n(n+1)}{2}}=2^{n(n+1)}$

Answer 2

Your mistake was writing $2^{n(n+1)}$ as$\tfrac124^{n(n+1)}$ rather than $4^{\tfrac12n(n+1)}$. The inductive step should be$$4^{\tfrac12n(n+1)}4^{n+1}=4^{\tfrac12n(n+1)+n+1}=4^{(\tfrac12n+1)(n+1)}=4^{(n+1)(n+2)}.$$