Question

$1+3+3^2+......+3^{n-1}=\frac{3^n-1}{2}$

Answer 1

Let P(n) : $1+3+3^2+.......+3^{n-1}=\frac{3^{n-1}}{2}$

For n = 1

p(1) : 1 = $\frac{3^1-1}{2}$

$1=1$

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

Let P(n) is true for n = k

$1+3+3^2+.....+3^{k-1}=\frac{3^k-1}{2}$ .......(1)

We have to show P(n) is true for n = k + 1

i.e., $1+3+3^2+......+3^k=\frac{3^{k+1}-1}{2}$

Now,

$\{1+3+3^2+.....+3^{k+1}\}+3^{k+1-1}=\frac{3k-1}{2}+3^k$ [Using equation (1)]

$=\frac{3^k-1+{2.3}^k}{2}$

$=\frac{{3.3}^k-1}{2}$

$=\frac{3^{k+1}-1}{2}$

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

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