Question

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}.$

Answer 1

Let P (n) $=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}$
For n = 1
$P\left(1\right)=\frac{1}{2^1}=1-\frac{1}{2^1}\Rightarrow \frac{1}{2}=frac12$
$\therefore$ P (1) is true
Let P (n) be true for n = k
$\therefore P\left(k\right)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}=1-\frac{1}{2^k}\text{For n = k + 1}P(k+1)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}+\frac{1}{2^{k+1}}=1-\frac{1}{2^{k+1}}=1-]frac12^k+\frac{1}{2^{k+1}}=1-(\frac{1}{2^k}-\frac{1}{2^{k+1}})=1-\left(\frac{2-1}{2^{k+1}}\right)=-\frac{1}{2^{k+1}}$
$\therefore$ P(k + 1 ) is true
thus P (k) is true $\Rightarrow P(k+1)$ is true
hence by principle fo mathematical induction,
p (n) is true for all $nϵN.$