I have these two finite geometric series:

$S_n$ = $\sum_{k=0}^n ar^k$

r$S_n$ = $\sum_{k=0}^n ar^{k+1}$

And then we substract both series so:

$S_n$ - r$S_n$ = a - $ar^{n+1}$ //this I understand

$\frac {S_n(1-r)}{1-r}$ = $ \frac{a - ar^{n+1}}{1-r}$ //where does 1-r come from?

  • 2
    $\begingroup$ $S_n-rS_n=(1-r)S_n$ $\endgroup$
    – Guest
    Dec 8 '15 at 20:52
  • $\begingroup$ @Guest could you elaborate, I don't get why they're equivalent $\endgroup$
    – boywonder
    Dec 8 '15 at 20:55
  • 1
    $\begingroup$ It's just the distributive property of $\cdot$ with respect to $+$ in $\mathbb{R}$ (more precisely, in this case, right distributivity of multiplication wrt addition). $\endgroup$
    – Guest
    Dec 8 '15 at 20:56

Once you have $$S_n - rS_n = a - ar^{n+1}$$ you can factor both sides: $$S_n ( 1 - r) = a(1 - r^{n+1})$$ (If that factorization isn't immediately obvious to you, just redistribute back through and confirm that you get the original expression again).

Now you can divide both sides by $(1-r)$ and get the desired conclusion.


In the second last step to the last step, you've divided both LHS and RHS by (1-r). In any equation, as long as you do the same operation to LHS and RHS it is allowed. for example - dividing both sides by a number (which is 1-r) in this case.

However, it would be easier to write the above system like this -

Sn-rSn=a-ar^(n+1) Thus, (1-r)Sn = a-ar^(n+1) Thus, Sn= [a-ar^(n+1)]/(1-r)

Hope this helps!


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