Question

1+3+3²+...+3n-1=(3n-1)/2

Answer 1

Answer:

$\red { 1+3+3^{2}+\cdot\cdot\cdot+3^{n-1}}$

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

Step-by-step explanation:

$Given \: 1 , 3 , 3^{2} , \cdot\cdot\cdot ,3^{n-1}$

$First \:term (a) = 1$

$\frac{a_{2}}{a_{1}} = \frac{3}{1} = 3$

$\frac{a_{3}}{a_{2}} = \frac{3^{2}}{3} = 3$

$\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = 3$

$Therefore, given \: sequence \:is \:in \:GP$

$Common \:ratio (r) = 3$

$Now , LHS = 1+3+3^{2}+\cdot\cdot\cdot+3^{n-1}\\= \frac{ 1(3^{n} - 1)}{3-1}$

$\boxed { \pink { Sum \:of \: n \: terms (S_{n}) = \frac{ a(r^{n} - 1)}{(r-1)} }}$

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

$= RHS$

$Hence ,\: Proved .$

•••♪