Manipulating expression $(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}})$ in inductive step.

Solution 1:

My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?

Note that we have to show that $$\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)\left(1-\frac{1}{2^{n+1}}\right)\color{red}{\ge}\frac 14+\frac{1}{2^{n+2}}$$

One has $$\begin{align}\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)\left(1-\frac{1}{2^{n+1}}\right)&= \frac 14-\frac{1}{2^{n+3}}+\frac{1}{2^{n+1}}-\frac{1}{2^{2n+2}}\\&= \frac 14+\frac{-2^{n-1}+2^{n+1}-1}{2^{2n+2}}\\&=\frac 14+\frac{3\cdot 2^{n-1}-1}{2^{2n+2}}\\&= \frac 14+\frac{2^n+(2^{n-1}-1)}{2^{2n+2}}\\&\color{red}{\ge}\frac 14+\frac{2^n+0}{2^{2n+2}}\\&=\frac 14+\frac{1}{2^{n+2}}\end{align}$$

Solution 2:

Note that you have to show that $$(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}})\geq \frac{1}{4}+\frac{1}{2^{(n+1)+1}}$$ which is $$\frac{1}{4}+\frac{1}{2^{n+1}}-\frac{1}{2^{n+3}}-\frac{1}{2^{2n+2}}\geq \frac{1}{4}+\frac{1}{2^{n+2}}$$ or $$\frac{1}{2^{n+1}}-\frac{1}{2^{n+3}}-\frac{1}{2^{2n+2}}\geq \frac{1}{2^{n+2}}.$$ Now multiply both sides by $2^{n+3}$, $$4-1-\frac{1}{2^{n+1}}\geq 2$$ or $$ 1\geq \frac{1}{2^{n+1}}$$ which holds.