Question

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

Answer 1

Let P(n) : $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{2^n}=1-\frac{1}{2^n}$

For n = 1

$\frac{1}{2}=1-\frac{1}{2^1}$

$\frac{1}{2}=\frac{1}{2}$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

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

Now,

$\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{2^k}\}+\frac{1}{2^{k+1}}$

$=1-\frac{1}{2^k}+\frac{1}{2^{k+1}}$

[Using equation (1)]

$=1-\left(\frac{2-1}{2^{k+1}}\right)$

$=1-\frac{1}{2^{k+1}}$

$\Rightarrow$ p(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI