Question

Prove the following by using the principle of mathematical induction for all $n\in N:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{2^n}=1-\frac{1}{2^n}$

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{2^n}=1-\frac{1}{2^n}$
For $n=1$, we have
$P\left(1\right)=\frac{1}{2}=1-\frac{1}{2^1}=\frac{1}{2}$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$, i.e.,
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{2^k}=1-\frac{1}{2^k}......\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider $(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{2^k})+\frac{1}{2^{k+1}}$
$=(1-\frac{1}{2^k})+\frac{1}{2^{k+1}}$
$=1-\frac{1}{2^k}+\frac{1}{{2.2}^k}$
$=1-\frac{1}{2^k}(1-\frac{1}{2})$
$=1-\frac{1}{2^k}\left(\frac{1}{2}\right)$
$=1-\frac{1}{2^{k+1}}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$