$a+ ar+ar^2 + ar^3 +ar^4+ \cdots+ ar^{n-1}=S_n$
$a\left(1+r+r^2 +r^3+\cdots+r^{n-1} \right) = S_n$
Iam trying to get $S_n = \dfrac{a(r^n -1)}{r-1}$
I don't know how to get $1+r+r^2 +r^3+\cdots+r^{n-1} =\dfrac{r^n -1}{r-1}$
Any help will be appreciated $:)$
Equation (1)
$S_n = a_1 + a_2 + a_3 + a_4 + …....... +a_n$
Putting value of each term,
Equation (2)
$S_n = a + ar + ar^2 + ar^3 + ar^4 + ... + ar^{n-1}$
Multiply equation (2) by r,
Equation (3)
$rS_n = ar + ar^2 + ar^3 + ar^4 + ........... + ar^{n}$
Subtract equation (2) from (3),
$rS_n - S_n = ar^n - a$
$S_n = \frac{a(r^n - 1)}{r - 1}$
Hint: $A= 1+r+r^2+...+r^n$, multiply with $r \implies rA=r+r^2+r^3+...+r^{n+1}$. Now, subtract $A-rA=1-r^{n+1} \implies A=\frac{1-r^{n+1}}{1-r}$ if $r\neq 1$.
Hints:
$$\begin{align}S_n-rS_n&=(a+ar+\dots+ar^n)-r(a+ar+\dots+ar^n)\\&=(\color{#4488dd}{a+ar+\dots+ar^n})-(\color{#44dd88}{ar+ar^2+\dots+ar^{n+1}})\\&=\color{#4488dd}a-\color{#44dd88}{ar}+\color{#4488dd}{ar}-\color{#44dd88}{ar^2}+\color{#4488dd}{ar^2}+\dots-\color{#44dd88}{ar^n}+\color{#4488dd}{ar^n}-\color{#44dd88}{ar^{n+1}}\\&=a-ar^{n+1}\end{align}$$