How do you find the sum of finite geometric series?

1 Answer
Bill K.
Jul 9, 2015

#a+ar+ar^2+\cdots+ar^n=\frac{a(r^{n+1}-1)}{r-1}# if #r!=1# and #a+a+a+\cdots+a=(n+1)a# when there are #n+1# terms (and #r=1#).

Explanation:

Consider a finite geometric series with #n+1# terms #a+ar+ar^2+\cdots+ar^n#. Call this sum #S_{n}#. Note that #rS_{n}=ar+ar^2+ar^3+\cdots+ar^{n+1}# so that #rS_{n}-S_{n}=ar^{n+1}-a#.

Solving this equation for #S_{n}# gives #S_{n}=\frac{a(r^{n+1}-1)}{r-1}# when #r!=1#. The number #a# is often referred to as the "first term" and the number #r# is often referred to as the "common ratio". The number #n# is the highest power of #r# in the sum and the sum itself has #n+1# terms.

If #r=1#, then #S_{n}=a+a+a+\cdots+a=(n+1)a#.

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