We're looking for the multiplicative inverse of $ 1+2^{p}+2^{n-p}$ mod $ 2^n$. Case 1: $ \frac{n}{p}$ is odd. The inverse is $ 1-2^p+2^{2p}-2^{3p}+\cdots -2^{n-2p}+2^n$ because \begin{align*}(1+2^p+2^{n-p})(1-2^p+2^{2p}-2^{3p}+\cdots -2^{n-2p}+2^n)&\equiv (1+2^p+2^{n-p})(1-2^p+2^{2p}-2^{3p}+\cdots -2^{n-2p})\\ &\equiv (1+2^p)(1-2^p+2^{2p}-2^{3p}+\cdots -2^{n-2p}) + 2^{n-p}\\ &\equiv 1-2^{n-p}+2^{n-p}\\ &\equiv 1\mod 2^n\end{align*} Case 2: $ \frac{n}{p}$ is even. The inverse is $ 1-2^p+2^{2p}-2^{3p}+\cdots+2^{n-2p}-2^{n-p+1}+2^n$ because $ (1+2^p+2^{n-p})(1-2^p+2^{2p}-2^{3p}+\cdots+2^{n-2p}-2^{n-p+1}+2^n)\\ \equiv (1+2^p+2^{n-p})(1-2^p+2^{2p}-2^{3p}+\cdots+2^{n-2p}-2^{n-p+1})\\ \equiv (1+2^p)(1-2^p+2^{2p}-2^{3p}+\cdots+2^{n-2p}-2^{n-p+1}) + 2^{n-p}\\ \equiv 1+2^{n-p} -2^{n-p+1} + 2^{n-p}\\ \equiv 1+2^{n-p+1}-2^{n-p+1}\\ \equiv 1\mod 2^n$ Now that we have found these inverses, we can use the fact that, for $ y>x$, $ 2^{y}-2^{x}=2^{x}+2^{x+1}+\cdots +2^{y-1}$ to write these inverses as the sums of powers of two directly.