Proof by induction of the volume of an n-dimensional box

Question

To solve a question I need a formula for the volume of an n-dimensional box with edges parallel to the coordinate axes centred at the origin. I am attempting to use induction. I am not entirely sure this is the correct way to approach. However, it is the only method I could think of. $$V(n)=\prod_{i=1}^n2x_i, n=2: V(2)=2^2x_1x_2$$ By observation we can see V(2) is true. Assuming the induction hypothesis that for an arbitrary k, the singular case n=k holds, meaning V(k) is true. $$V(k)=\prod_{i=1}^k2x_i=2^k(x_1x_2...x_k)$$ It follows that: $$V(k+1)=2^{k+1}(x_1x_2...x_kx_{k+1})=2^k(x_1x_2...x_k)2x_{k+1}$$ I am unsure of how to proceed here. I do not believe I can finish by saying $V(k+1)=2V(k)x_{k+1}$.

Answer 1

The volume of an $n$-dimensional rectangular box $B =[a_1,b_1] \times \dots \times [a_n,b_n]$ is defined as $$V(B) =\prod_{i=1}^n (b_i -a_i) .$$

In your case you have $a_i = -x_i, b_i = x_i$. Thus $b_i - a_i = 2x_i$ and $$V = \prod_{i=1}^n (2x_i) = \left(\prod_{i=1}^n 2\right)\left( \prod_{i=1}^n x_i\right) = 2^n \prod_{i=1}^n x_i.$$