Sum of singular 1-cubes = boundary of a singular 2-cube?

[gothloli]

Homework Statement

I'm doing question 23 in Chapter 4 of Spivak's Calculus on Manifolds. The question asks,
For R > O, and n an integer, define the singular l-cube,
$c_{R,n} :[0,1] \rightarrow \mathbb {R}^2 - 0$ by $c_{R,n} (t) = (Rcos2\pi nt, Rsin2\pi nt).$ Show that there is a singular 2-cube $c:[0,1]\rightarrow \mathbb {R}^2 - 0$ such that $c_{R_1,n} - c_{R_2,n} = \partial c$

Homework Equations

$\partial c = \sum_{i =1}^{n} \sum_{\alpha = 0,1} (-1)^{i+ \alpha} c_{i, \alpha}$
$c_{i, \alpha} = c(I^n_{i,\alpha})$

The Attempt at a Solution

What I'm confused about is probably the notation of
$c_{R_2,n}$ and it's relation with $c_{(i, \alpha)}$, are they the same thing since
$c_{R,n} (t) = (Rcos2\pi nt, Rsin2\pi nt)$
, which I find difficult to connect with the
$(i,\alpha)$
-face of c and I.
To solve the question you use $\partial$ c, to get the (i, n) - face of c, from which c can be obtained. The solution is apparently
$c(t_1,t_2) = (t_1R_1 + (1-t_1)R_2)(cos2\pi nt_2, sin2\pi nt_1)$

 

[lippyka]

But I am not sure how to get this from the given equation c_{R_1,n} - c_{R_2,n} = \partial c. Any help would be greatly appreciated!