Prove that sum, Sn, of first n term geometric series...

Is given by Sn= a(1-r^n)/(1-r). I know Sn = a +ar + ar^2 .... but I don't know what to do from there....

Is given by Sn= a(1-r^n)/(1-r). I know Sn = a +ar + ar^2 .... but I don't know what to do from there....

S_n = a +ar + ar^2+\cdots+ar^{n-1}

rS_n = ar +ar^2 + ar^3+\cdots+ar^n

What's

S_n-rS_n

(edited 8 years ago)

Sn = a +ar + ar^2+...+ar^{n-1}

rSn = ar +ar^2 + ar^3+...+ar^n

What's Sn-rSn?

Why rSn?

Why rSn?

Because multiplying by r shifts the index by one unit. See below.

Spoiler

Because multiplying by r shifts the index by one unit. See below.

Spoiler

Thanks, I'm assuming you just have to remember this or will you be expected to answer something similar in the exam such as with the sum of infinity?

Thanks, I'm assuming you just have to remember this or will you be expected to answer something similar in the exam such as with the sum of infinity?

The sum to infinity is

\displaystyle \sum_{k=0}^{\infty}r^k = \lim_{n \to \infty}\sum_{k=0}^{n}r^k = \lim_{n \to \infty} \frac{1-r^{n+1}}{1-r} = \frac{1}{1-r}.

\alpha

is any real number we have

\displaystyle \alpha \sum_{k=0}^{\infty}r^k = \sum_{k=0}^{\infty}\alpha r^k = \frac{\alpha}{1-r}.

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